lecture 4- stoichiometry

7/24/2019 Lecture 4- Stoichiometry

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Stoichiometry

andRate Laws

Suchithra Thangalazhy Gopakumar

H83RED Reactor Design

Autumn Semester 2015


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Outline

Rate Law and Reaction Rate Constant

Arrhenius Law

Elementary reaction

Order of Reaction

Rate Law for reversible reactions Stoichiometric Table

Batch systems:

Constant volume and

Constant pressure

Flow systems:

Liquid phase

Gas phase reactions that have a change in the total number of moles

Reactors with variable volumetric flowrates


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Chemical kinetics deals with reaction rates

Types of Reactions

Homogeneous reaction

involves substances of only one phase

Heterogeneous reaction

involves more than one phase, and thereaction usually occurs at or very near the

interface between the phases

Phase

Irreversible reaction

proceeds in only one direction and continues

until the reactants are exhausted

Reversible reaction

proceeds in both forward and reverse

directions; continues until the equilibrium state

is reached

Equilibrium

A B

A B

Types of Reactions


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The Rate Law and the Reaction Rate Constant

General form of the rate law, which is an algebraic equation representing the

reaction rate as a function of the concentrations (activities) of various species:

(1)

k = reaction rate constant

~ independent of the species concentrations

,...)C,C(f)T(kr BAA

The specific reaction rate

constant, kA, is always referred to

a particular species

Exception

OHNaClHClNaOH kkkkk 2

NaOH + HCl NaCl + H2O

Liquid Phase:

Reaction rate constantis an intensive parameter concentration-independent

Gas Phase:

k=f(temperature, total pressure, ionic strength, choice of solvent )

k=f(temperature, catalysts, total pressure )

Rate Law


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A = pre-exponential factor or frequency factorE= activation energy (J/mol or cal/mol),

R = gas constant (8.314 J/mol.K=1.987 cal/mol.K)

T = absolute temperature (K)

(2)

Activation energy, E characterise energy state of molecules which could

result in their chemical interaction

Minimum energy that molecules must have to react

(3)

Temperature Dependency from Arrhenius Law

RT/EAe)T(k

1/T, K-1

ln

k,sec-1

slope= -E/R

1/T, K-1

ln

k,sec-1

Low E

High E

TR

EAkA

1lnlnActivation energy from experimental results

E , k~ sensitive to the temperature

E , k~ less sensitive to thetemperature

Rate Law


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Temperature rise needed to double the rate of reaction

Average

TemperatureActivation Energy, E

40 kJ/mol 160 kJ/mol 280 kJ/mol 400 kJ/mol

0o

C 11o

C 2.7o

C 1.5o

C 1.1o

C400 oC 65 oC 16 oC 9.3 oC 6.5 oC

1000 oC 233 oC 58 oC 33 oC 23 oC

Reactions with high activation energies are very temperature-sensitive

Any given reaction is more temperature-sensitive at lower temperatures thanat higher temperatures

Rules of thumb:

Rate of reaction doubles for every 10o

Cincrease in temperature

~ Only true for a specific combination

of activation energy and temperature1/T, K-1

ln k, sec-1

Low E

High E

Arrhenius Law


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Task

Consider the following elementary reactions :

Solut ion

1) Which reaction has the higher activation energy?

2) Which reactions have the same activation energy?

3) Which reaction is virtually temperature insensitive?

Arrhenius Law


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Solut ion

4) Which reaction will dominate (i.e. take place the fastest) at high temperatures?

5) Which reaction will take place the fastest at low temperatures?

Arrhenius Law


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Finding k(T) given k(T0)

0/0)( RTEAeTk

RTEAeTk /)(

k(T)k(T0)exp E

R

1

T0

1

T

If we know:

The specific reaction rate k0(T

0) at a temperature T

0 The activation energy, E,

We can find the specific reaction rate constant k(T) at any other

temperature,

T, for that reaction

0/

0RTEeTkA substitute

RT

E

RT

E

eeTkTk

0)()( 0

Arrhenius Law


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Task

Determine the activation energy, E and frequency factor, A from the following data:

Solut ion

Arrhenius equation in logarithm form:

k (min-1) 0.001 0.050

T (oC) 0.0 100.0

TRE

AkA1

lnln

RT/EAe)T(k

Arrhenius Law


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Elementary and Non- elementary Reactions

Assumption from Collision theory:

Reactant particles must collide with each other.

Reaction is a result of the collision of a single molecule A with a singlemolecule B

E: Energy required to activate the molecules into a state from whichreactant bonds can change into product bonds.

Only those collisions with enough energy to exceed E can lead toreaction.

Rate law can be derived from collision theory

(1),...)C,C(f)T(kr BAA

Rate of reaction is proportional to the number of such collisions.

The number of collisions is proportional to the concentration ofreactant.

Elementary Reaction


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The dependence of possible collisions on the reactant concentrations.

A

A

B

B

A

A

B

B

A

4 collisions

Add anothermolecule of A

6 collisions

Add anothermolecule of B

A

A

B

B

A B

9 collisions

Elementary Reaction


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Elementary Reactions

corresponds to the stoichiometric

equationBAA CkCr

If kinetic experiments results in a rate law, which corresponds tothe stoichiometric equation, we call such reactions elementary.

Molecularity Elementary step Rate law

1. Unimolecular A -> products rate = k CA

2. Bimolecular2A -> productsA + B -> products

rate = k C2Arate = k C

AC

B

3. Termolecular3 A -> productsA + 2 B -> productsA + B + C -> products

rate = k C3Arate = k CAC

2B

rate = k CACBCC

A + B R

The molecularity of an elementary reaction is the number of molecules

(atoms) involved (colliding) in the reaction.

Elementary Reaction


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Rate law reflects a limiting step or combination of limiting steps

The rate limiting step is the slowest step in the sequence of stepsleading to product formation

The overall rate of reaction depends on this slowest step

Non-elementary reactions involve more complex, multi-step reactions

H2 + I2 2HIElementary Reaction

222 IHH CkCr

H2 + Br2 2HBr

2

22

BrHBr2

21BrH1

HBrC/Ck

CCkr

/

Non-ElementaryReaction

Elementary and Non-Elementary Reactions

Elementary Reaction


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The Reaction Order and Rate Law

General form of the rate law: (4)

2 NO + O2 2 NO2

...CCkr BAAA

Temperature-dependentterm

Concentration-dependentterms

The order of a reaction refers to the power to which the concentrations are rais

The overall order of the reaction: n = + +

2O

2

NONONO CCkr

CO + Cl2 COCl2

23CO

/ClCO CkCr

H2 + Br2 2HBr

2

22

BrHBr2

21

BrH1

HBr

C/Ck

CCkr

/

2N2O 2N2 + O2

2

22

2

O

ONON

ON1 Ck

Ckr

2OCk1>> Pseudo-first order with

respect to nitrous oxide

2OCk1


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Examples of Forms of Rate Law and Units of Reaction Rate Constant

Zero-order:AA kr

secdm

mol

secdm

mol33 AA

kr secdm

mol3

k

First-order:AAA Ckr

[k]=sec-1

secdm

mol

dm

mol

sec

1

secdm

mol333 AAA Ckr

Second-order: 2

AAA Ckr

secmol

dm3k

secdm

mol

dm

mol

secmol

dm

secdm

mol36

22

3

3 AAA Ckr

Third-order: 3

AAA Ckr

secmol

dm2

6

k

secdm

mol

dm

mol

secmol

dm

secdm

mol39

33

2

6

3 AAA Ckr

Rate Law


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TaskWhat is the reaction rate law for the reaction:if the reaction is elementary?Calculate the rates of A, B, and C in a CSTR where the concentrationsareCA = 1.5 mol/dm

3, CB = 9 mol/dm3 and kA = 2 (dm

3/mol)(1/2)(1/s).

C B2

1A

Rate Law


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(6)

One more limit condition for reversible reactions: Equilibrium State

The rate of reaction is identical for all species and equal zero

(5)

Rate Law for Reversible Reactions

aA + bB k

cC + dDk-1

At thermodynamic equilibrium:

b

Be

a

Ae

d

De

c

Ce

CCC

CCK

KC is an equilibrium constant Units

abcd

3

dm

mol

The equilibrium constant characterise: equilibrium concentration of all species

ratio between rates of forward andreverse reactions

1-reverse

forward

k

k

k

k

KC

-rA 0

k= Forward reaction rate constantk-1 = Reverse reaction rate constant

The state in which the concentrations of the reactants and productshave no net change over time

Rate Law (Reversible Reaction)


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Kinetic rate law for reversiblereaction must be consistentwith

thermodynamicequilibrium relationship

additional limit conditionwith specified rate law offorward and reversereactions

AkA

Bk-A

An elementary reversible reaction

(7)

formation of species A (8)

Reaction rate with respect to species A

AAA Ckr disappearance of species A

AAA Ckr

Reaction rate with respect to species A

BAA Ckr (9)formation of species A

b

Be

a

Ae

d

De

c

CeC

CCCCK

Forward reaction

Reverse reaction

A BkA

ABk-A



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The overall rate of formation ofspecies A

(10)BAAA

reverseAforwardAA

CkCk

rrr

,,

(11)

The overall rate of disappearance of species A:

)( BA

AAABAAA

reverseAforwardAA

C

k

kCkCkCk

rrr

,,

(12)

Rate law for elementary monomolecular reversible reaction:

)(C

BAAA

K

CCkr

Check of the consistency with thermodynamic equilibrium:

)(0C

BeAeAA

KCCkr

After rearranging:C

BeAe

K

CC

Ae

Be

C C

CK

Ae

Be

C C

CK

Thermodynamics tells that:

At equilibrium:

AkA

Bk-A

1-reverse

forward

k

k

k

kKC



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TaskWrite the rate law for the elementary reaction

Here kfA and krA are the forward and reverse specific reaction rates bothdefined with respect to A.



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Stoichiometry provides information on relative quantities of reactants

and products in chemical reactions. We have already addressed related topics such as limiting reactant

and excess in our previous lecture.

Stoichiometry

Stoichiometry

Rate of reaction ri is a variable in all the design equations for

reactors. Rate of reaction is a function of thermodynamic quantities

such as temperature and also concentration of the species.

Conversion of the limiting reactant is a function of the concentrationand therefore rate of reaction is a function of the conversion: ri=f(Xi).

To evaluate the integrals associated with the design equations ofthe form

the functional form of ri(X) has to be found.

Stoichiometry provides a solution for this challenge.

)( jA Cfr )(XhC jj )(XgrA

rate law From a stoichiometric table


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Mole Balance for Different Reactors

Reactor Type Differential Algebraic Integral

Batch

CSTR

PFR

PBR

Vrdt

dXN AA 0

)(

0

0

tX

A

A

A

Vr

dXNt

exit

0

)( A

A

r

XFV

A

A r

dV

dXF0

ou t

in

X

X A

Ar

dXFV 0

A

A rdW

dXF0

ou t

in

X

X A

Ar

dXFW 0

exit)r(

XC

A

A

0

ou t

in

X

X A

Ar

dXC 0

Conversion: Revision of concepts


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Relative quantities (extent)

Stoichiometry: Extent

+ + 0 0 0 0

0 0 0 0

Reaction

Initial quantities

Quantities remaining at

t=t

Quantities reacted

Amount of B react with A = 0

0 = 0 0

= 0

0 =

0 =

0 =

0 =

Therefore the relative quantities are given by the following equation

[13]

The extent E of the reaction is the relative number of moles reacted/ generat


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Relative reaction rates

Stoichiometry: Relative rates

+ + Reaction Rates

Rate of disappearance of A =

Rate of disappearance of B = Rate of appearance of C =

=

=

=

Therefore the relative reaction rates are given by

[14]


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Constructing stoichiometric table

Stoichiometry: Stoichiometric table

Consider the reaction

t=0

NA0NB0NC0ND0NI0

t=t

NANBNCNDNI

+ +

I stands for inert substances, e.g.solventsReaction was terminated at time t when

the conversion of the limiting reactantA

reachedX. Composition of the product:

NA, NB, NC, ND and NI moles of substances

A, B, C, D and I, respectivelyFrom the definition of conversion

= 0 0

Xreachedconversion

whenreacted

Aofmoles

volumesystem

theintointroduced

Aofmoles

Xconversionat

volumesystema

inAofmoles

For every mole of A reacted, b/a moles of B must react

)(reactedmolesreactedBmoles 0XNa

bA

a

bA


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Species Initially

(mol)

Change

(mol)

Remaining

(mol)

A NA0

B NB0

C NC0

D ND0

I NI 0

Total NT0

)( 0XNA

)( 0XNab A

)( 0XNa

cA

)( 0XN

a

dA

XNNN AAA 00

XNabNN ABB 00

XNa

cNN ACC 00

XN

a

dNN ADD 00

XNa

b

a

c

a

dNN ATT 00 1

0II NN

A + b/a B c/a C + d/a D

XNa

d

a

c

a

bA01

)(

0

0

tX

A

A

A

Vr

dXNt

i i i i i


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Total number of moles changed


The total number of moles of all active species (reactants and products) at

conversionX of speciesA:

= 0+ +

1 0

The change in the total number of moles per mole of A

reacted

=+

1 =0 0

= 0+ 0 where

= + = /

0 = 0+ 0

We will introduce concentrations in thestoichiometric table using:

=

[15]

[16]

St i hi t St i hi t i t bl


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In order to define all concentrations on a basis per mole of A we

introduce the parameter :

The final equation for concentration of species B:

0

0

0

0

0

0

A

i

A

i

A

ii

y

y

C

C

N

N

V

XN)a/b(N

V

XN)a/b(NC AABABB

0000

V

Xa/bNC BAB

)(0 [18]

Expressing all in terms of NA0

We now need Vas a function of conversionX in order to obtain )(XhC jj

[17]



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= 00

()

Initially

(mol)

Change

(mol)

Remaining

(mol) (mol/dm3)

A NA0

B

C

D

I

Total NT0

)( 0XNA

)( 0XNa

bA

)( 0XNa

cA

)( 0XNad

A

XNNN AAA 00

Xa

bNN BAB 0

Xa

cNN CAC 0

X

adNN DAD 0

0II NN

XNNN ATT 00

V

X1NC 0AA

)(

V

NC II

0

V

Xa/bNC BAB

)(0

V

Xa/cNC CAC

)(0

V

Xa/dNC DAD)(0


00 ABB NN

00 ACC NN

00 ADD NN

00 AII NN



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Constant-Volume (or Density) Reaction Systems

Sealed vessel (gas phase reaction)

~Example:

Laboratory bomb calorimeter

(to measure the heat of combustion

of a particular reaction)

Isothermal gas-phase reaction with no change in the total number of moles

(ideal gas law applies)

~ Example: coal gasification

Liquid phase reaction in an incompressible solvent

(except polymerizations)

~ Solvent dominates

~ Changes in the density of the solute do not affect

the overall density of the solution significantly

CO + H2O CO2 + H2P

T

nRV

nRTPV

http://upload.wikimedia.org/wikipedia/commons/3/34/Bombenkalorimeter_mit_bombe.jpghttp://upload.wikimedia.org/wikipedia/commons/3/34/Bombenkalorimeter_mit_bombe.jpg


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= 0

0

()

Initially

(mol)

Change

(mol)

Remaining

(mol) (mol/dm3)

A NA0

B NB0

C NC0

D ND0

I NI 0

Total NT0

)( 0XNA

)( 0XNa

bA

)( 0XNa

cA

)( 0XNad

A

XNNN AAA 00

Xa

bNN BAB 0

Xa

cNN CAC 0

X

adNN DAD 0

0II NN

XNNN ATT 00

)1(0 XCC AA

XabCC BAB )( /0

Xa/cCC CAC )(0

XadCC DAD )/(0

0II CC

Stoichiometric Table for a Batch, Constant-Volume System




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)( XCC AA 10


BAA CkCr

XabCXkCr BAAA )( /).1( 00

XabCC BAB )( /0

XabXkCr BAA )( /)1(02 )(XfrA

exit)(

0

A

A

r

XFV

Rate law

Design Equation ou t

in

X

X A

Ar

dXFV 0

Sizing of the batch, CSTR, PFR system

St i hi t i T bl E l


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Task

A component of soap (sodium stearate) is produced by saponification of glyceryl

stearate with aqueous caustic soda according the following equation:

3 NaOH + (C17H35COO)3C3H5 3 C17H35COONa + C3H5(OH)3

Production is carried out in a batch reactor. The initial mixture consists solely of sodium

hydroxide at a concentration of 10 mol/dm3 and glyceryl stearate at a concentration

of 2 mol/dm3. What is the concentration of glycerine when the conversion of sodium

hydroxide is (a) 20% and (b) 90%?

Stoichiometric Table: Example

St i hi t i T bl Fl t


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Stoichiometric Table for Flow Systems

EnteringFA0FB0FC0FD0FI0

LeavingFAFBFCFDFI


Batch System

)1(000 XNXNNN AAAA

[19]

Concentration of each species in a stream:

volume

moles

timevolume

timemoles

q

FC

j

j

/

/

Flow systems

)X(FXFFF AAAA 1000

Nj0

Nj

Fj0

Fj

In flow systems we deal with molar

fluxes

Replaced by

Stoichiometric Table: Flow system

St i hi t i T bl Fl t


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Concentrations of species A, B, C, and D in terms of entering molarflow rate (FA, FB, FC, FD), the conversion X, and the volumetric flowrate

[20]

)1(0 Xq

F

q

FC AAA

q

XFabF

q

FC ABBB

00 )/(

q

XFacF

q

FC ACCC

00 )/(q

XFadF

q

FC ADDD

00 )/(

0

0

0

0

0

0

0

0

A

i

A

i

A

i

A

ii

y

y

C

C

vC

vC

F

F

The parameter :

Expressing all in terms of FA0 + +




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Species Initially(mol / time) Change(mol / time) Remaining(mol / time)

A FA0

B

C

D

I

Total FT0

00 ABB FF

00 ACC FF

00 ADD FF

00 AII FF

)( 0XFA

)( 0XF

a

bA

)( 0XFa

cA

)( 0XFa

dA

)1(0 XFF AA

Xa

bFF BAB 0

Xa

cFF CAC 0

Xa

dFF DAD 0

IAI FF 0

XFa

b

a

c

a

dFF ATT 00 1

XFFF ATT 00

Stoichiometric Table for Flow Systems




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For liquid-phase reactions, the volumetric flow rate does not practically

change in the course of reaction: v =v0 = Const

)1(0 XFF AA

X

a

bFF BAB 0

Xa

cFF CAC 0

Xa

dFF DAD 0

Stoichiometric Table for Flow Systems (Liquid-Phase Reactions)

)1()1( 00

0 XCXq

FC A

AA

X

a

bCC BAB 0

Xa

cCC CAC 0

Xa

dCC DAD 0

Therefore for a given rate law we have:)(Xgr

A


Stoichiometric Table


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For most batch and liquid-phase reactions and some gas-phase reactions

the volumetric flow rate or reaction volume does not change, however:

Volume Change with Reaction

volume (or volumetric flow rate) will change in the following situations:

Flow system with gas-phase reaction with a change in the total

number of moles

Flow system with non-isothermal gas-phase reaction

Batch system with variable volume

01a

b

a

c

a

d


+ +

N2 + 3H2 2NH3



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Equation of State is a fundamental equation for gas systems:

RTZNPV T [21]Z = a compressibility factorNT= an overall number of moles in a gas mixture

[22]

By dividing equation (21) by equation (22):

00000 T

T

N

N

Z

Z

VP

PV

T

T

Volume Change with Reaction

The first four columns of the stoichiometric tables are always valid

Assumptions concerning a volume change were made when expressing

concentration in terms of conversion

Expressing volume as a function of conversion V = f (X) (Batch Systems)

000

00

T

T

N

N

Z

Z

P

PVV

T

T [23]

00000

RTNZVPT

time = 0




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Total number of moles in system as a function of

conversion:

yA0 is the molar fraction of

species A in an initial

mixture

[24]

[25]

000

0

0T

T

N

N

Z

Z

P

PVV

T

T

XNNN ATT 00

XyXN

N

N

NA

T

A

T

T

0

0

0

0

11

Let the parameter :

[26]

reactorthetofedmolesofnumbertotal

conversioncompleteformolesofnumberin totalchange

00

0

0

0

1 AT

A

T

A

yN

N

N

N

a

b

a

c

a

d

Equation [25] now becomes:

XXy

N

NA

T

T 11 00

Equation [23]:

)1(

00

00 X

T

T

Z

Z

P

PVV [28]

Volume Change with Reaction in a Batch System


[27]



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In gas-phase systems, at the certain ranges of temperatures and pressuresthe compressibility factor does not change significantly during the course of

the reaction: Z0Z

[29]

The volume of gas mixture for a variable volume batch reactorat anytime/conversion:

)1(00

0

0 XTT

ZZ

PPVV

0

00 )1(

T

TX

P

PVV

Volume Change with Reaction in a Batch System

If the reactor is a rigid steel containerof constant volume, then:

0VV

and equation [29] can be used to calculate the pressure change as afunction of TandX




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Consider the total concentration at

any point within the reactor:

[30]ZRT

P

qZRT

Pq

q

FC TT

[31]

At the entrance to the

reactor:

00

0

0RTZ

PCT

Taking the ratio of the equations(31) and (30) and assuming that

Z0Z:

[32]0

0

0

0T

T

P

P

F

Fqq

T

T

XXyX

F

F

F

FA

T

A

T

T 111 00

0

0

Total molar flux rate as a

function of conversion in acontinuous reactor:

[33]XFFF ATT 00

[34]

Volumetric flow rate as a functionof conversion can be obtained bycombining the equations (32) and

(34) :

0

00 )1(

T

T

P

PXqq [35]

Volume Change with Reaction in a Flow System




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If the volumetric flow rate is known as a function of conversion, then theconcentration of any species in a flow system can be defined as a functionof conversion:

Thus:

[36]

0

00 )1(

T

T

P

PXqq

q

FC

j

j

Molar flow rate of each species as a function of conversion is:

XSFF jjAj 0

T

T

P

P

Xq

XSFC

jjA

j 0

00

0

)1(

)(

T

T

P

P

X

XSCC

jjA

j0

0

0

)1(

)(

Concentrations as a Function of Conversion




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T

T

P

P

X

XsCC

jjA

j0

0

0

)1(

)(

Initially

(mol/time)

Change

(mol/time)

Remaining

(mol/time) (mol/volume. time)

A FA0

B

C

D

I

Stoichiometric Table for a Variable-Volume Gas Flow System

T

T

P

P

X

XCC AA

0

0

0)1(

1

T

T

P

P

X

Xa/bCC BAB

0

0

0)1(

)(

T

T

P

P

X

Xa/c

CC

C

AC

0

00 )1(

)(

T

T

P

P

X

Xa/dCC DAD

0

0

0)1(

)(

T

T

P

P

X

CC IAI

0

0

0

)1(

IAI FF 0

Xa

dFF DAD 0

Xa

cFF

CAC 0

Xa

bFF BAB 0

)1(0 XFF AA

00 ABB FF

00 ACC FF

00 ADD FF

00 AII FF

)( 0XFA

)( 0XFa

bA

)( 0XFa

cA

)( 0XFa

dA


+ +

Stoichiometric Table: Flow system-Example


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Task

A mixture of 28% SO2 and 72% air is charged to a flow reactor in which SO2 is

oxidized.

2 SO2 + O2 2 SO3

Calculate concentrations of all species at the following conversion of SO2: 0; 0.25;

0.5; 0.75; 1, if the total pressure is 1485 kPa and the temperature is constant at

227oC. Note that air composition is 21% O2 and 79% N2.

Solution

Taking SO2 as the basis of calculations:

+12

Stoichiometric Table: Flow system Example

2 SO2 + O2 2 SO3

Stoichiometric Table: Flow system-Example


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Task

The elementary gas phase reaction

3A + 2B

3 C +5 D

is carried out in a flow reactor operated isothermally at 427C and 28.7 atmospheres.

Pressure drop can be neglected. Express the rate law and the concentration of each

species as a function of conversion and as a function of the total molar flow rates. The

entering volumetric flow rate is 10 dm3/s and the specific reaction rate,k is 200

dm12/mol4s. The feed is equal molar in A and B.

Solut ion

k1

A is the limiting reactant

Stoichiometric Table: Flow system Example

Volume change (Flow system) -Example


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Task


3A + 2B

3 C +5 D




entering volumetric flow rate is 10 dm3/s and the specific reaction rate is 200 dm12/mol4s.

The feed is equal molar in A and B.

k1

Volume change (Flow system) Example

Volume change (Flow system) -Example


49/50

Task


3A + 2B

3 C +5 D




entering volumetric flow rate is 10 dm3/s and the specific reaction rate is 200 dm12/mol4s.

The feed is equal molar in A and B.

k1

Volume change (Flow system) Example

Summary


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Summary

lecture 4- stoichiometry

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