ERT 316: REACTION ENGINEERING CHAPTER 3 ... - …portal.unimap.edu.my/portal/page/portal30/Lecturer...

ERT 316: REACTION ENGINEERING

CHAPTER 3

RATE LAWS &

STOICHIOMETRY

1

OUTLINE

PART 1: Rate Laws

Relative Rates of Reaction

Reaction Order & Rate Law

Reaction Rate Constant, k

PART 2: Stoichiometry

Batch System Stoichiometric Table

Flow System Stoichiometric Table

Calculation for Concentration in terms of

Conversion

1. RELATIVE RATES OF REACTION

d

r

c

r

b

r

a

r DCBA

dDcCbBaA

212

22 NOONOrrr

EXAMPLE

Reaction

Stoichiometry

If NO2 formed at 4 mol/m3/s

(r NO2= 4 mol/m3/s), what is

the rate of formation of NO?

22 22 NOONO


2

//4

2

3 smmolrNO

212

22 NOONOrrr

smmolsmmol

rNO //42

//42 3

3

22

2NONOrr

If NO2 formed at 4 mol/m3/s (r NO2= 4 mol/m3/s),

what is the rate of formation of NO?

22 22 NOONO


The Reaction:

is carried out in a reactor. If at a particular point,

the rate of disappearance of A is 10 mol/dm3/s,

what are the rates of B and C?

CBA 532

EXERCISE


The relative rates are

Given, the rate of disappearance of A, -rA, is 10mol/dm3/s

Thus, solving the rates of B & C;

532

CBA rrr

CBA 532

r A= -10 mol/dm3/s

sdmmolrB //102

3 3

32

BA rr

52

CA rr

sdmmolrC //102

5 3

sdmmol //15 3 sdmmol //25 3

2. REACTION ORDER & RATE LAW

The reaction rate (rate of disappearance) depends

on temperature and composition.

It can be written as the product of reaction rate

constant, kA and a function of concentrations

(activities) of the reactants involved in the

reaction:

..., BAAA CCfnTkr

Rate law is a kinetic

expression that gives the

relationship between

reaction rate, -rA, and

concentration.

2. REACTION ORDER & RATE LAW

For reaction in which the stoichiometric coefficient is 1

for ALL species:

we shall delete the subscript on the specific reaction

rate, (e.g.; A in kA) to let

OHNaClHClNaOH kkkkk2

Rate law is a kinetic

expression that gives the

relationship between

reaction rate, -rA, and

concentration.

OHNaClHClNaOH 21111

2.1 POWER LAW MODELS & ELEMENTARY RATE LAWS

Power Law Model:

The rxn is 𝛂 order wrt reactant A

AND

The rxn is 𝛃 order wrt reactant B

The overall order of the reaction, n;

BAA CkCr

n


The unit of the specific reaction, k, will vary with the

order of reaction.

Products

A

Time

ionConcentratk

n

1

AA kr sdmmolk 3/Zero order (n=0)

First order (n=1) AAA Ckr 1 sk

Second order (n=2) 2

AAA Ckr smoldmk /3

Third order (n=3) 3

AAA Ckr 123 / smoldmk


Elementary reaction: a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.

Elementary rate law:

The rxn is said to follow the elementary rate law if the stoichiometic coefficients are IDENTICAL to the reaction order of each species.

A Products

BA Products

Unimolecular reaction

Bimolecular reaction

22 22 NOONO

2

2

ONONONO CCkr

EXAMPLES OF REACTION RATE LAWS

Non-elementary rate laws: reactions that do not follow simple rate laws (power rate laws).

Example 1: Homogeneous Rxn

The kinetic rate law is:

Rxn order: first order wrt to CO, three-halves order wrt Cl2, five-halves order overall.

23

2ClCOCO CkCr

2.2 NON-ELEMENTARY RATE LAWS

22 COClClCO

Gas phase

synthesis of

phosgene

Example 2: Heterogeneous Rxn

The rate of disappearance of toluene per mass of

catalyst is :

where KB & KT is the adsorption constants.

TTBB

TH

TPKPK

PkPr

1

2'

2.2 NON-ELEMENTARY RATE LAWS

4662356 CHHCHCHHCcat

Gas-solid catalyzed

rxn:

Hydrodemethylation

of toluene (T)

In terms of partial

pressure rather than

concentrations

MBHTcat

2

follows Langmuir-

Hinshelwood kinetics

2.3 REVERSIBLE REACTIONS

For reversible rxn, all rate laws must reduce to the

thermodynamic relationship relating the reacting

species concentrations at equilibrium.

dDcCbBaA ⇌

b

eB

a

eA

d

eD

c

CeC

CC

CCK

Thermodynamic

Equilibrium

Relationship


21012662 HHCHC ⇌

2

, BBforwardB Ckr

EXAMPLE: combination rxn of 2 mol of benzene to form 1 mol

H2 and 1 mol diphenyl.

kB

k-B

22 HDB ⇌ kB

k-B

symbolically;

The rate of disappearance of benzene; 2

, BBforwardB Ckr

The reverse rxn btween diphenyl & hydrogen;

6621012 2 HCHHC ⇌ k-B

OR

2, HDBreverseB CCkr

The rate of formation of benzene (in reverse direction);


The net rate of formation of benzene is;

Multiplying both sides by -1, we obtain the rate law of

disappearance of benzene, -rB

reverseBforwardBnetBB rrrr ,,,

2

2

HDBBB CCkCk

2

2

HDBBBB CCkCkr

2

2

HD

B

BBB CC

k

kCk


Replacing the ratio of the reverse & forward rate law

constant by equilibrium constants;

where

C

B

B Kk

k

2

2

HD

B

BBBB CC

k

kCkr

C

HD

BBBK

CCCkr 22

Concentration

equilibrium constant

3. THE REACTION RATE CONSTANT

RTE

A AeTk /A= preexponential factor or frequency factor

E= activation energy, J/mol or cal/mol

R=gas constant = 8.314 J/mol-K = 1.987 cal/mol-K

T= absolute temperature, K

Arrhenius

equation

-no of collision

RTEe /

A

-probability that

the collision will

result in a reaction


RTE

A AeTk / Activation energy is a measure of the minimum

energy that the reacting molecules must have

in order for the reaction to occur (energy required to

reach transition state).

Reactants Products

Transition state

Energy barier -total no of collision

RTEe /

A

probability that

- the collision will

result in a rxn

k - no of collision that

result in a rxn


RTE

A AeTk /

Taking a natural logarithm;

E ⬆, k ⬆, -r = ⬆

The larger the

activation energy,

the more

temperature

sensitive k and thus

the reaction rate.

TR

EAkA

1lnln

4. BATCH SYSTEMS STOICHIOMETRIC TABLE

Purpose of developing stoichiometric table:

To determine the no of moles of each species

remaining at a conversion of X.


Species Initially

(mol)

Change

(mol)

Remaining

(mol)

A

B

C

D

I

Totals

Components of stoichiometric table:

refers to moles of

species reacted or

formed

Recall from Chapter 2:

Factorizing;


0

0

A

AA

N

NNX

XNNN AAA 00

XNN AA 10

moles of A reacted

aA + bB cC + dD

moles of A remaining

in the reactor at a

conversion

of X


Moles B

reacted, NB

XNa

dA0

XNa

cA0

Moles C

formed, NC

Moles D

formed, ND

XNa

bA0

Moles B reacted

Moles A reacted

Moles A reacted


moles B remaining

in the system, NB

XNa

bN AB 00

NC

moles of B

initially in the

system

moles of C

formed

XNa

cN AC 00

ND XN

a

dN AD 00

moles of B

reacted

moles of D

formed


Species Initially

(mol)

Change

(mol)

Remaining (mol)

A

B

C

D

I -

Totals

0AN

0BN

0CN

0DN

0IN

XNa

bA0

XNa

cA0

XNa

dA0

XNA0

XNa

cNN ACC 00

XNa

dNN ADD 00

XNa

bNN ABB 00

XNNN AAA 00

0II NN

XNa

b

a

c

a

dNN ATT 00 1

0TN


XNa

b

a

c

a

dNN ATT 00 1

Total no of moles per mole of A reacted can be

calculated as:

where

XNN AT 00

1a

b

a

c

a

d

Change in the total number of moles per

mole of A reacted


Species Initially Change Remaining Concentration

A

B

C

D

I

Totals

Can we express concentration of each species??

AA kCr 2

AA kCr 3

AAA Ckr

Concentration of each species in terms of

conversion can be expressed as:


V

XN

V

NC AA

A

10

V

XNabN

V

NC ABB

B00 /

Remaining (mol)

A

B

C

D

XNa

cNN ACC 00

XNa

dNN ADD 00

XNa

bNN ABB 00

XNNN AAA 00

V

XNacN

V

NC ACC

C00 /

V

XNadN

V

NC ADD

D00 /

Recall from

stoichiometric

table

V

XabNNN ABA // 000


V

XNabNC AB

B00 /

V

XNacNC AC

C00 /

V

XacNNN ACA // 000

V

XabN BA /0

V

XacN CA /0

V

XadNNN ADA // 000


V

XNadNC AD

D00 /

V

XadN DA /0

0

0

0

0

0

0

A

i

A

i

A

ii

y

y

C

C

N

N



A

B

C

D

I -

0AN

0BN

0CN

0DN

0IN

XNa

bA0

XNa

cA0

XNa

dA0

XNA0

XNa

cNN ACC 00

XNa

dNN ADD 00

XNa

bNN ABB 00

XNNN AAA 00

0II NN

V

XNC A

A

10

V

XabNC BA

B

/0

V

XacNC CA

C

/0

V

XadNC DA

D

/0

IOC

X

a

b

N

NNXN

a

bNN

A

BAABB

0

0000

X

a

bN BA0

0

0

0

0

0

0

A

i

A

i

A

ii

y

y

C

C

N

N



A

B

C

D

I -

0AN

0BN

0CN

0DN

0IN

XNa

bA0

XNa

cA0

XNa

dA0

XNA0

X

a

bNN BAB 0

XNNN AAA 00

0II NN

V

XNC A

A

10

V

XabNC BA

B

/0

V

XacNC CA

C

/0

V

XadNC DA

D

/0

IOC

X

a

cNN CAC 0

X

a

dNN DAD 0

Given the saponification for the formation of soap

from aqueous caustic soda & glyceryl stearate is:

Letting X the conversion of sodium hydroxide, set up a

stoichiometric table expressing the concentration of

each species in terms of its initial concentration and the

conversion.


3533517533517 33 OHHCCOONaHCHCOOHCaqNaOH

EXAMPLE


3533517533517 33 OHHCCOONaHCHCOOHCaqNaOH

We know that this is a liquid-phase reaction.

Therefore, V=V0

XC

V

XN

V

XNC A

AAA

1

110

0

00

DCBA 33

XC

V

XabNC BA

BAB

3

1/0

0

0

1313 dcba

EXAMPLE



A

B

C

D

I -

Total 0

0AN

0BN

0CN

0DN

0IN

XNA03

1

XNA0

XNA03

1

XNA0

XNN BAB

3

10

XNN AA 10

0II NN

XCC AA 10

XCC BAB

3

10

XCC CAC 0

XCC DAD

3

10

IOC

XNN CAC 0

XNN DAD

3

10

EXAMPLE

0TN 0TT NN

5. FLOW SYSTEMS STOICHIOMETRIC TABLE

Purpose of developing stoichiometric table:

To determine the effluent flow rate of each species at

a conversion of X.


Species Feed rate to

reactor

(mol/time)

Change within

the reactor

(mol/time)

Effluent rate

from reactor

(mol/time)

A

B

C

D

I

Totals

Components of stoichiometric table:


Species Feed rate

to reactor

(mol/time)

Change

within the

reactor

(mol/time)

Effluent rate from

reactor (mol/time)

Concentration

(mol/L)

A

B

C

D

I -

Totals

0AF

00 ABB FF

00 ACC FF

00 ADD FF

00 AiI FF

XFa

bA0

XFa

cA0

XFa

dA0

XFA0

X

a

bFF BAB 0

XFFF AAA 00

IAI FF 0

X

a

cFF CAC 0

X

a

dFF DAD 0

0TF XFFF ATT 00

XFC A

A

10

XabFC BA

B

/0

XacFC CA

C

/0

XadFC DA

D

/0

IA

I

FC

0

QUIZ 3

Given a liquid phase reaction:

A+ 2B C + D

The initial concentration of A and B are 1.8 kmol/m3

and 6.6 kmol/m3 respectively. Construct a

stoichiometric table for a flow system considering A as

the basis of calculation.

ANSWER FOR QUIZ 3

A+ 2B C + D

Given:

From stoichiometry, we know that,

3

3

0

/6.6

/8.1

mkmolC

mkmolC

BO

A

0

0

0

0

0

0

A

i

A

i

A

ii

y

y

C

C

F

F

67.38.1

6.6B

1121 dcba

0

0

A

ii

C

C

08.1

0C

3

3

0

/0

/0

mkmolC

mkmolC

DO

C

Since C & D are

products.

08.1

0D

11 a

b

a

c

a

d

ANSWER FOR QUIZ 5


reactor

(mol/time)

Change

within the

reactor

(mol/time)

Effluent rate from

reactor (mol/time)

A

B

C

D

Totals

0AF

BAB FF 00

CAC FF 00

DAD FF 00

XFA02

XFA0

XFA0

XFA0

XFF BAB 20

XFF AA 10

XFF AC 0

XFF AD 0

0TF XFFF ATT 00

ANSWER FOR QUIZ 5


reactor

(mol/time)

Change

within the

reactor

(mol/time)

Effluent rate from

reactor (mol/time)

A

B

C

D

Totals

0AF

00 67.3 AB FF

00 CF

00 DF

XFA02

XFA0

XFA0

XFA0

XFF AB 267.30

XFF AA 10

XFF AC 0

XFF AD 0

0TF XFFF ATT 00

Substituting the numerical values;

1. For liquid phase:

Batch System:

6. CONCENTRATION IN TERMS OF CONVERSION

0VV

V

XN

V

NC AA

A

10

V

NC B

B

V

NC C

C

V

NC D

D

V

XabN BA /0

V

XacN CA /0

V

XadN DA /0

0

0 /

V

XabN BA

0

0 /

V

XacN CA

0

0 /

V

XadN DA

XabC BA /0

XacC CA /0

XadC DA /0

1. For liquid phase:

Flow System -


0

XFFC AA

A

10

B

B

FC

C

C

FC

D

D

FC

XabF BA /0

XacF CA /0

XadF DA /0

0

0 /

XabF BA

0

0 /

XacF CA

XabC BA /0

XacC CA /0

XadC DA /0

0

0 /

XadF DA

2. For gas phase:

Batch System

From equation of state;

At any time t,

At initial condition (t=0)


V

NC A

A Need to substitute V

from gas law equation

RTZNPV T

T= temperature, K

P= total pressure, atm (1 atm= 101.3 kPa)

Z= compressibility factor

R= gas constant = 0.08206 dm3-atm/mol-K

00000 RTNZVP T

(1)

(2)

2. For gas phase:

Batch System

Dividing (1) by (2);


RTZNPV T

00000 RTNZVP T

(1)

(2)

000

00

T

T

N

N

Z

Z

T

T

P

PVV

Recall from stoichiometric table

XNNN ATT 00 (4)

Dividing (4) by NT0 ;

XN

N

N

N

T

A

T

T 0

0

0

1

XyA01

(3)

2. For gas phase:

Batch System

Applies for both

batch and flow

systems


XyN

NA

T

T0

0

1

XN

N

T

T 10

0

01T

A

N

N

a

b

a

c

a

d

0Ay

At complete conversion (for irreversible rxn): X=1, NT=NTf

XN

NN

T

TT

0

0

Rearranging;

0

0

T

TTf

N

NN

Will be substitute

in (3)

2. For gas phase:

Batch System

Substituting the expression for NT/NT0 in (3),


000

00

T

T

N

N

Z

Z

T

T

P

PVV

(3)

XZ

Z

T

T

P

PVV

1

00

00

If the compressibility factor are not change

significantly during rxn, Z0⩳Z

0

00 1

T

TX

P

PVV

(5)

2. For gas phase:

Flow System

From gas law, at any point in the reactor,

At the entrance of reactor;


ZRT

PFC T

T

0

0

0

0T

T

P

P

F

F

T

T (3)

00

0

0

00

RTZ

PFC T

T

(1)

(2)

Dividing (1) by (2)

j

j

FC

Need to substitute υ


0

0

0

0T

T

P

P

F

F

T

T

2. For gas phase:

Flow System

Substituting for FT;


Recall from stoichiometric table

XFFF ATT 00

0

0

0

000

T

T

P

P

F

XFF

T

AT

0

0

0

00 1

T

T

P

PX

F

F

T

A

0

000 1

T

T

P

PXyA

0

00 1

T

T

P

PX (4)

2. For gas phase:

Flow System

Substituting υ & Fj;

0

00 1

T

T

P

PX


(4)

j

j

FC

Need to substitute υ


0

00

0

1T

T

P

Px

XvFC

jjA

j

XvFF jjjj 0(5)

T

T

P

P

x

XvC

jj

A0

0

01

Stoichiometric

coefficient

(d/a, c/a, -b/a, -a)

0

00 1

T

T

P

PX

2. For gas phase:

Flow System

Concentration for each species:


aA + bB cC + dD

XFFC AA

A

10

B

B

FC

C

C

FC

D

D

FC

XabF BA /0

XacF CA /0

XadF DA /0

0

00

1

/

P

P

T

T

x

XabC B

A

0

00

1

/

P

P

T

T

x

XacC C

A

0

00

1

/

P

P

T

T

x

XadC D

A

0

00

1

1

P

P

T

T

x

XCA

I

I

FC

IIF

0

0

00

1 P

P

T

T

x

C IA

SUMMARY

Relative rate of reaction:

Power Law Model:

d

r

c

r

b

r

a

r DCBA

dDcCbBaA

BAA CkCr

SUMMARY

Elementary rate law:

The rxn that in which its stoichiometic coefficients are IDENTICAL to the

reaction order of each species.

Non-elementary rate laws:

The reactions that do not follow simple rate laws (power rate laws) in

which its stoichiometic coefficients are NOT IDENTICAL to the reaction

order of each species.

Reversible reaction:

All rate laws must reduce to the thermodynamic relationship relating the

reacting species concentrations at equilibrium.

Power Law Model:

SUMMARY

Reaction Rate Constant, k

RTE

A AeTk /

E ⬆, k ⬆, -r ⬆

The larger the activation energy,

the more sensitive k is, (towards

the change in temperature)

SUMMARY

Stoichiometric Table for Batch Systems

Species Initially Change Remaining

A

B

C

D

I -

0AN

0BN

0CN

0DN

0IN

XNa

bA0

XNa

cA0

XNa

dA0

XNA0

XNa

cNN ACC 00

XNa

dNN ADD 00

XNa

bNN ABB 00

XNNN AAA 00

0II NN


reactor

(mol/time)

Change within the

reactor (mol/time)

Effluent rate from

reactor (mol/time)

A

B

C

D

I -

Totals

SUMMARY

Stoichiometric Table for Flow Systems

0AF

00 ABB FF

00 ACC FF

00 ADD FF

00 AiI FF

XFa

bA0

XFa

cA0

XFa

dA0

XFA0

X

a

bFF BAB 0

XFFF AAA 00

IAI FF 0

X

a

cFF CAC 0

X

a

dFF DAD 0

0TF XFFF ATT 00

Expression of V and υ in calculating the concentration of each species:

Batch systems

Liquid phase:

Gas phase:

Flow systems

Liquid phase:

Gas phase:

SUMMARY

0VV

0

P

P

T

TXVV 0

0

0 1

P

P

T

TX 0

0

0 1

EXERCISE

Given a gas phase reaction:

A+ 2B 2C

Construct a stoichiometric table for a flow system

considering A as the basis of calculation with

equimolar feed of 50% A and 50% B. Derive the

concentration for each species (for an isothermal

gas phase reaction, with no pressure drop).

.

Spe

cies

Feed rate

to reactor

(mol/time)

Change

within the

reactor

(mol/time)

Effluent rate

from reactor

(mol/time)

A

B

C

0AF

00 AB FF

00 CF

XFA02

XFA02

XFA0

XFF AB 210

XFFF AAA 00

XFF AC 02

XFFC AA

A

10

B

B

FC

C

C

FC

XFA0

0

00

5.01

21

P

P

T

T

X

XCA

0

00

5.01

2

P

P

T

T

X

XCA

0

00

5.01

1

P

P

T

T

X

XCA

XFA 210

From stoichiometry table

0

00

0

0

0RT

Py

RT

PC AA

A

ERT 316: REACTION ENGINEERING CHAPTER 3 ... - …portal.unimap.edu.my/portal/page/portal30/Lecturer...

Documents

Transcript of ERT 316: REACTION ENGINEERING CHAPTER 3 ... - …portal.unimap.edu.my/portal/page/portal30/Lecturer...