Chemical Reaction Engineering

Lecture 2

General algorithm of Chemical Reaction Engineering

• Mole balance• Mole balance

• Rate laws• Rate laws

• Stoichiometry• Stoichiometry

• Energy balance• Energy balance

•Combine and Solve•Combine and Solve

Classification of reactions

• Phases involved:– Homogeneous reaction – reaction that occur in one phase – Heterogeneous reaction – reaction that involves more

than one phase and usually occurs at the interface between the phases (e.g. heterogeneous catalysis)

• Equilibrium position– Reversible reaction – reaction that can proceed in either

direction depending on the concentration of reagents and products

– Irreversible reaction – reaction that at given conditions can be assumed to proceed in one direction only (i.e. reaction equilibrium involves much smaller concentration of the reagents)

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)

• Probability of meeting 3 molecules is very small, so uni-molecular and bimolecular reaction are the only two to consider

Elementary reactions

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

= =

Relative Rates of Reaction

• if we are interested in species A we can use A as the basis of calculation and define the reaction rate with respect to A

aA bB cC dD+ ⎯⎯→ +

b c dA B C Da a a

+ ⎯⎯→ +

• 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

,...)],()][([ BAAA CCfnTkr =−

dr

cr

br

ar DCBA ==

−=

−

Rates of Reaction

• units of the reaction rate constant

,...)],()][([ BAAA CCfnTkr =−

timeionConcentratk

n−

=1)(][

algebraic function of concentrations

Arrhenius law/( ) E RTk T Ae−=

...⋅⋅ βαBA CC

• reaction order: ...++= βαn

• reaction rate is found experimentally, data on frequency factor (A), activation energy (E) and the order of the reaction can be found in relevant handbooks.

Non-Elementary Reaction rates

• The reaction rate dependence on the concentration and temperature will become more complicated when a reaction comprising several elementary steps is considered (incl. catalytic and reversible reactions)

Reversible reaction• 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 −= = − +

B D

Equilibrium constant

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−=

in terms of concentration⎟⎟

⎠

⎞⎜⎜⎝

⎛−=−

c

HDBB K

CCCkr 22

12

• reaction rates in terms of other reagents:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−=

c

HDB

BD

KCC

Ckrr 22121

Few more facts of K• From thermodynamics 0ln rRT K G=−Δ

• for gases, K can be defined in terms of pressures or concentrations

/ /

/

c a d aC D

p b aA B

p pKp p

=/ /

/ ( ) ; 1c a d aC D

C pb aA B

C C c d bK K RTC C a a a

δ δ−= = = + − −

• temperature dependence:

2

( )ln rH Td KdT RT

Δ=

0

1 21

( ) 1 1( ) ( )exp rp p

H TK T K TRT T T

⎡ ⎤⎛ ⎞Δ= −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

0

2

( ) ( )ln r R p RH T c T Td KdT RT

Δ +Δ −=

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⎛ ⎞

− ⋅ −⎜ ⎟⎝ ⎠=

Summary of the Reactor design equationsIn the previous lecture we found design equations for various reactors:

To solve them we need to find disappearence rate as a function of conversion:

)(XgrA =−

Relative Rates of Reaction

• 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 =

• Let’s see how we can relate it to the reaction rate for various types of reactors.

Batch reactorb c dA B C Da a a

+ ⎯⎯→ +

• conversion:Moles of A reacted

Moles 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:

δ

δ - the total molar increase per mole of A reacted XNNN ATT ⋅⋅+= 00 δ

Batch reactor• Now, if we know the number of moles of every component

we can calculate concentration as a function of conversion. ( )

( )( )

( )( )

( )( )V

XadNVNC

VXacN

VNC

VXabN

VXNabN

VNC

VXN

VNC

DADD

CACC

BAABBB

AAA

+Θ==

+Θ==

−Θ=

−==

−==

0

0

000

0 1

• However, generally V can be a function of X as well...• In a constant volume reactor (e.g. batch reactor, liquid

reactor): ( )( )( ) etc. ,;1

0

0

XabCCXCC

BAB

AA

−Θ=−=

Flow reactors

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

Flow reactors

• Stoichiometric table for a flow system

Flow reactors

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

• 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 −⎜ ⎟⎝ ⎠

timelitertimemoles

vFC A

A ==

Reactions involving volume change• In a gas phase reaction a molar flow rate might

change as the reaction progresses

2 2 33 2N H NH⎯⎯→+ ←⎯⎯4 moles 2 moles

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

Batch reactor with variable volume• As such it would be a rare case (e.g. internal combustion

engine), but a good model case:

• If we divide the gas equation at any moment in time by the one at moment zero:

RTZNPV T=compressibility factor

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

000

00

T

T

NN

ZZ

TT

PPVV

XXNN

NN

T

A

T

T εδ +=⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛11

0

0

0

( )XTT

PPVV ε+⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛= 1

0

00 function V=g(X)

Flow reactor with variable flow rate

TT

F PCv ZRT

= =

• Using the gas equation we can derive the total concentration as:

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

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

=1

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 3.6 (p.118)

• For reaction above 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.

Concentration equilibrium constant: 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 =

equilibrium conversion:

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

⎡ ⎤− = − −⎢ ⎥

⎣ ⎦

– PFR – in the next lecture ☺

40.4

30

| 7 10

1714dm|

A X

A

A X

rF XVr

−=− = ⋅

= =−

Problems

• Class: P3-15• Home: P3-7, P3-13