Chemical Reaction Engineering

Lecture 6

Elementary reactions• Kinetics of chemical reactions determined by the elementary reaction steps.• Molecularity of an elementary reaction is the number of molecules coming

together to react in one reaction step (e.g. uni-molecular, bimolecular, termolecular)

Uni-molecular: first order in the reactant

[ ] [ ]d AA P k Adt

⎯⎯→ = −

Bimolecular: first order in the reactant

[ ] [ ][ ]d AA B P k A Bdt

+ ⎯⎯→ = −

Proportional to collision rate

2H Br HBr Br+ ⎯⎯→ +

! rate of disappearance of individual components can be calculated as:

1 i

i

dn dvdt dt

ξν

= =

Equilibrium constant• Let’s consider a reaction of diphenyl formation

1

16 6 12 10 22 k

kC H C H H

−

⎯⎯→ +←⎯⎯

• rate of change for benzene

2

26 61 1

[ ] 2 2B B D Hd C H r k C k C C

dt −= = − +

• noticing that the equilibrium constant: 1 1CK k k−=

226 61

[ ] 2 D HB B

C

C Cd C H r k Cdt K

⎡ ⎤= = − +⎢ ⎥

⎣ ⎦in terms of concentration

! At equilibrium rb=0, so 2 2220;D H D H

B CC B

C C C CC K

K C⎡ ⎤

+ = =⎢ ⎥⎣ ⎦

Reaction rate• Reaction rate is not a constant, it depends

– on concentration of the reagents– on the temperature– on the total pressure in the reactions

involving gas phase– ionic strength and solvent in liquid state– presence of a catalyst

226 61

[ ] 2 D HB B

C

C Cd C H r k Cdt K

⎡ ⎤= = − +⎢ ⎥

⎣ ⎦

Le Chatelier Principle

Dependence on the concentraion can be calculated knowing reaction mechanism, as before

Temperature dependence: Arrhenius equation

/( ) E RTk T Ae−=

Reaction rate

226 61

[ ] 2 D HB B

C

C Cd C H r k Cdt K

⎡ ⎤= = − +⎢ ⎥

⎣ ⎦

Dependence on the concentration can be calculated knowing reaction mechanism, as before

Temperature dependence: Arrhenius equation

/( ) E RTk T Ae−=

• Below, we will consider only the concentration and temperature dependence of the reaction rate

Reaction rate• From the collision theory, only molecules

with the energy higher than the activation energy can react:

• The usual “rule of thumb that the reaction rate doubles with every 10ºC is not always true:

1 2

1 1/

1 2( ) ( )E R

T Tk T k T e⎛ ⎞

− ⋅ −⎜ ⎟⎝ ⎠=

Back to the reactors

• if we are interested in species A we can define A as the basis of calculation

aA bB cC dD+ ⎯⎯→ +

b c dA B C Da a a

+ ⎯⎯→ +

• conversion: Moles of A reactedMoles of A fedAX =

• number of moles A left after conversion:

( )0 0 0 1A A A AN N N X N X= − = −

• number of moles B left after conversion:

0 0B B AbN N N Xa

= −

( )0 1AAA

N XNCV V

−= =

( ) ( )( )00 0 A BB ABB

N b a XN b a N XNCV V V

Θ −−= = =

Batch reactors• For every component in the reactor we can write after

conversion X is achieved:

d

Flow reactors

• Equations for flow reactors are the same with number of moles N changed for flow rate F [mol/s].

Flow reactors

• For a flow system a concentration at any point can be obtained from molar flow rate F and volumetric flow rate

moles/time molesliters/time liter

AA

FCv

= = =

• For reaction in liquids, the volume change is negligible (if no phase change occurred):

( )0 1AA A

FC C Xv

= = − 0B A BbC C Xa

⎛ ⎞= Θ −⎜ ⎟⎝ ⎠

Flow reactor with variable flow rate• In a gas phase reaction a molar flow rate might

change as the reaction progresses2 2 33 2N H NH⎯⎯→+ ←⎯⎯4 moles 2 moles

TT

F PCv ZRT

= =

• In a gas phase reaction a molar flow rate might change as the reaction progresses

0 00

0 0

TT

F PCv Z RT

= =at the entrance:total concentration:

00

0 0 0

T

T

PF Z Tv vF P Z T

=total volume rate:

0 000 0

0 0 0 0

j jTj j T

T T

F F Z TPPF Z TC F v CF P Z Tv F P Z T

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

Flow reactor with variable flow rate• In a gas phase reaction a molar flow rate might

change as the reaction progresses

( )( )

0

00

0

1

A j jj

F XC P Tv X

P T

ν

ε

Θ +=

+

0T T AF F F Xδ= + ⋅ ⋅

00

0 0 0

T

T

PF Z Tv vF P Z T

= ( ) ( )0 00 0 0

0 0

1 1AP PT Tv v y X v XP T P T

δ ε= + = +

Example

• For reaction below calculate– equilibrium conversion in a constant batch reactor;– equilibrium conversion in a flow reactor;– assuming the reaction is elementary, express the rate of

the reaction– plot Levenspil plot and determine CSTR volume for 80%

conversion• assume the feed is pure N2O4 at 340K and 202.6kPa.

KC=0.1mol/l; kA=0.5 min-1.

2 4 22N O NO⎯⎯→←⎯⎯

Example1. Batch reactor

( )2 2 2 2

0 0

0

4 41 1

Be A e A eC

Ae A e e

C C X C XKC C X X

= = =− −

30 00

0

0.071mol/dm .AA

y PCRT

= =

0.44eX =

Example2. Flow reactor

( ) ( )( )2 2 2 2

0 0

0

4 41 1 1

Be A e A eC

Ae A e e e

C C X C XKC C X X Xε

= = =− − +

30 00

0

0.071mol/dm .AA

y PCRT

= =

( )( )

( )0 0

0

0

1 11 1

1

A AAA

AB

F X C XFCv v X XC XC

X

ε ε

ε

− −= = =

+ +

=+

0.51eX =

Example3. Rate laws 2

BA A A

C

Cr k CK

⎡ ⎤− = −⎢ ⎥

⎣ ⎦

– Constant volume ( )2 2

00

41 AA A A

C

C Xr k C XK

⎡ ⎤− = − −⎢ ⎥

⎣ ⎦

– Flow( )

( )

2 20 0

2

1 41 1A A

A AC

C X C Xr kX K Xε ε

⎡ ⎤−− = −⎢ ⎥

+ +⎢ ⎥⎣ ⎦

– Levenspiel plot

Example4. CSTR volume for X=0.4, feed of 3 mol/min

– Constant volume ( )2 2

00

41 AA A A

C

C Xr k C XK

⎡ ⎤− = − −⎢ ⎥

⎣ ⎦

– Flow

40.4

30

| 7 10

1714dm|

A X

A

A X

rF XVr

−=− = ⋅

= =−

Example: Hippo’s digestion

• A hippo's stomach consists of three distinct chambers - the parietal blind sac, the stomach, and the glandular stomach which connects to the intestines.

CSTRFermentation reaction

PFR

Hippo’s digestion

• Assuming Monod kinetics in the stomach:A M

A M

kC CrC K

− =+

A M P M+ ⎯⎯→ +

( )0 01 ; ( )A A B A B MC C X C C Y X= − = Θ +

20

10

2

21

(1 )( )(1 )

1 1;1

A M MAM

A M

AM

kC X Y XrC X K

b cX eX aXaX r b cX eX

− +Θ− = =

− +

+ − − −= =

− + −

• Stomach (CSTR)

Hippo’s digestion

• Assuming Michealis-Menten kinetics:A

A M

kCrC K

− =+

A E EA P E+ ⎯⎯→ ⎯⎯→ +

( )0 1 ;A AC C X= −

2

1 1 (1 )(1 )AM

b Xr a X− + −

=−

• Intestine (PFR)

Hippo’s stomach• Maximum conversion rate in the stomach

• Maximum conversion rate in the intestine:

• CSTR is more efficient for autocatalytic reaction; PFR is more efficient for enzyme catalyzed reaction

Hippo’s stomach• for CSTR

• for PFR

• studies indicate that X2=45%; 75% of it happens in CSTR

Hippo’s stomach• Our starting point:

X0=0; X1=0.34 X2=0.45;

• Fitting the data

• flow rate 40 kg/day, with density of grass 365kg/m3, volumetric rate 0.13 m3/day

V1=0.45m3

Questions• A hippo is full grown in 4.5 years. A baby hippo follows its mother around

and eats one half of what she eats. However, the rate of digestion in the autocatalytic CSTR stomach, -rAM1, is the same as the mother hippo as is the conversion. Assuming that in the growing stages the hippo stomach volume is proportianal to its age. How old is the baby hippo?

• What is the conversion in the intestine (PFR) for the baby hippo diiscussedin Question (1) assuming that Vmax is one half that of the mother?

• The hippo has picked up a river fungus and now the effective volume of the CSTR stomach compartment is only 0.2 m3. The hippo needs 30% conversion to survive. Will the hippo survive?

• The hippo had to have surgery to remove a blockage. Unfortunately, the surgeon, Dr. No, accidentally reversed the CSTR and PFR during the operation. Oops!! What will be the conversion with the new digestive arrangement? Can the hippo survive?