Chemical Reaction Engineering 1Chemical Reaction Engineering 1

제제 33 장장Rate LawsRate Laws

and Stoichiometryand Stoichiometry

반응공학 반응공학 11

Booker T. Washington

“Success is measured not so

much by the position one has

reached in life, as by the

obstacles one has overcome

while trying to succeed”

An African-American Scholar and Inventor

After completing Chapter 3, reader will be able to:

Write a rate law and define reaction order and activation energy.

Set up a stoichiometric table for both batch and flow systems and express concentration as a function or conversion.

Calculate the equilibrium conversion for both gas and liquid phase reactions.

Write the combined mole balance and rate law in measures other than conversion.

Set up a stoichiometric table for reactions with phase change.

Objectives

- Homogeneous reaction is one that involves only one phase

- Heterogeneous reaction involves more than one phase. (reaction usually occurs at or very near the interface between the phase)

- Irreversible reaction is one that proceeds in only one direction and continues in that direction until the reactants are exhausted.

- Reversible reaction can proceed in either direction, depending on the concentration of reactants and products relative to the corresponding equilibrium concentration.

Basic Definitions

Basic Definitions

- Molecularity of a reaction is the number of atoms, ions or molecules involved in a reaction step.

- The term, unimolecular, bimolecular, and termolecular refer to reactions involving, respectively, one, two, or three atoms (or molecules) interacting in any one reaction step.

NaBrOHCHBrCHNaOH

HeThU

HIIH

NOONO

33

42

23490

23892

22

22

2

22

Example of Elementary Reaction

Relative Rates of Reactions

If the rate law depends on more than one species, we MUST relate the concentrations of different species to each other. A stoichiometric table presents the stoichiometric relationships between reacting molecules for a single reaction.

In formulating our stoichiometric table, we shall take species A as our basis of calculation (i.e., limiting reactant) and then divide through by the stoichiometric coefficient of A. In order to put everything on a basis of “per mole of A.”

Da

dC

a

cB

a

bA

aA + bB cC + dD

d

r

c

r

b

r

a

r DCBA

(3-1)

(2-1)

(2-2)

The relationship can be expressed directly from the stoichiometry of the reaction.

Relative Rates of Reactions

Example:

2NO + O2 2NO2

rNO=

rO2=

rNO2

-2 -1 2

If rNO2= 4 mol/m3-sec rNO= -4 mol/m3-sec

rO2= -2 mol/m3-sec

The dependence of the reaction rate –rA on the concentration of the species is almost without exception determined by experimental observation.

BAAA CCkr (3-3)

order with respect to reactant A order with respect to reactant Bn (=) : the overall order of the reaction

The order of a reaction refers to the powers to which the concentrations are raised in the kinetic rate law.

Reaction Order and Rate Law

Unit of Specific Reaction Rate

A products

The unit of the specific reaction rate, kA, vary with the order of the reaction.

s

moldmkCkrorder-Third

smol

dmkCkrorder-Second

skCkrorder-First

sdm

molkkrorder-Zero

AAA

AAA

AAA

AA

233

32

3

)/(}{:

)(}{:

1}{:

)(}{:

k=(Concentration)1-n

Time

(3-4)

(3-5)

(3-6)

(3-7)

2

2ONONONO CCkr

nd order w.r.t. nitric oxide

1st order w.r.t. oxygen

overall is 3rd order reaction

2NO + O2 2NO2

2/3

2ClCOCO CkCr

CO + Cl2 COCl2

Kinetic rate raw

In general, first- and second-order reactions are more commonly observed.

Elementary reaction Non-elementary reaction

1st order w.r.t. carbon monoxide

3/2 order w.r.t. chorine

overall is 5/2 order reaction

Elementary and Nonelementary Reaction

Reversible Reactions

All rate raws for for reversible reactions must reduce to the thermodynamic relationship relating the reacting species concentrations at equilibrium. At equilibrium, the rate of reaction is identically zero for all species (i.e., -rA=0). For the general reaction

aA + bB cC + dD

The concentrations at equilibrium are related by the thermodynamic relationship

])/[( 3 abcdbBe

aAe

dDe

cCe

C dmmolCC

CCK

RT

E

A AeTk

)(

Specific reaction rate (constant)

Arrhenius frequency factoror

pre-exponential factor mathematical numbere=2.71828…

Activation energy, J/mol or cal/mol

Gas constant8.314 J/mol · K1.987 cal/mol · K8.314 kPa · dm3/mol · K

Absolute Temperature, K

Arrhenius equation

RT

E

A AeTk

)(

Arrhenius Equation

T 0, kA 0T , kA A

A

kA

T (K)

Activation energyActivation energy E : a minimum energy that must be possessed by reacting molecules before the reaction will occur.

RT

E

e

The fraction of the collisions between molecules that together have this minimum energy E (the kinetic theory of gases)

Activation energy E is determined experimentally by carrying out the reaction at several different temperature.

TR

EAkA

1lnln

Activation energy

A+BC A-B-C AB+C

Example 3-1: Determination of Activation energy

ClCl

N=N

Decomposition of benzene diazonium chloride to give chlorobenzene and N2

T(K) k (s- 1)313 0.00043319 0.00103323 0.00180328 0.00355333 0.00717

Calculate the activation energy using following information for this first-order

1/T (K- 1) k (s- 1)0.003195 0.000430.003135 0.001030.003096 0.001800.003049 0.003550.003003 0.00717

TR

EAkA

1lnln

Finding the activation energyPlot (ln k) vs (1/T)

[Solution 1]

Select two data points, (1/T1, k1) and (1/T2, k2), and then calculate.

Therefore, E 116 kJ/mole

[Solution 2]

Find a best fit from data point

ln k=14,017/T + 37.12

Slope=E/R=14,017E=116.5 kJ/mole

TR

EAkA

1lnln 1

1

TR

EAkA

1lnln 2

2

ln k2

k1

= - ( )ER ( )1

T2

1T1

-

임의의 온도에서 속도상수를 구할 수 있다 .

Finding Activation EnergyPlot (ln k) vs (1/T)

0.01

0.001

0.00010.0030 0.0031 0.0032

1/T (K-1)

k (s

ec-1)

0.005

0.0005

0.0030250.00319

ln k=14,017/T + 37.12

Activation Energy

The larger the activation energy, the

more temperature-sensitive is the

rate of reaction.

Typical values of 1st order gas-phase

reaction

Frequency factor ~ 1013 s-1

Activation energy ~ 300 kJ/mol

1/T (K-1)

0.01

0.001

0.00010.0030 0.0031 0.0032

k (s

ec-1)

High E

Low E

E 를 알고 나면 실험하지 않은 어느 온도에서의 k 값을 구할 수 있다 .

Specific Reaction Rate

TTR

Ekk

Aek

Aek

RT

E

RT

E

11exp

00

00

Taking the ratio

Present Status of Our ApproachDesign Equations

Vrdt

dXN AA 0

Batch

CSTR

PFR

PBR

AA rdV

dXF 0

AA rdW

dXF 0

A

A

r

XFV

0

X

AA Vr

dXNt

00

X

AA r

dXFV

00

X

AA r

dXFW

00

Differentialform

Algebraicform

Integralform

Stoichiometry

Da

dC

a

cB

a

bA

aA + bB cC + dD

d

r

c

r

b

r

a

r DCBA

(3-1)

(2-1)

(2-2)

The relationship can be expressed directly from the stoichiometry of the reaction.

d

r

c

r

b

r

a

r DCBA

(3-1)

Batch System

t = tNA

NB

NC

ND

Ni

t = 0NA0

NB0

NC0

ND0

NI0

Da

dC

a

cB

a

bA

At time t=0, we will open the reactor and place a number of moles of species A, B, C, D, and I (NA0, NB0, NC0, ND0, and NI0, respectively)

Species A is our basis of calculation.

NA0 is the number of moles of A initially present in the reactor.

NA0X moles of A are consumed as a result of the chemical reaction.

NA0-NA0X moles of A leave in the system.

The number of moles of A remaining in the reactor after conversion X

NA = NA0 - NA0X = NA0(1-X)

Stoichiometric Table for a Batch System

00

0000

0000

0000

0000

)(

)(

)(

)(

)(

(mol) (mol) (mol)

Remaining Change Initially Species

III

ADDAD

ACCAC

ABBAB

AAAAA

NNNinertI

XNa

dNNXN

a

dND

XNa

cNNXN

a

cNC

XNa

bNNXN

a

bNB

XNNNXNNA

XNa

b

a

c

a

dNNNTotal ATTT 000 1

Concentration of Each Species

V

XadN

V

XNadN

V

NC

V

XacN

V

XNacN

V

NC

V

XabN

V

XNabN

V

NC

V

XN

V

NC

y

y

C

C

N

N

DAADDD

CAACCC

BAABBB

AAA

A

i

A

i

A

ii

)/()/(

)/()/(

)/()/(

)1(

000

000

000

0

0

0

0

0

0

0

Constant Volume Batch Reactor

Flow Systems

Da

dC

a

cB

a

bA

liter

moles

eliters/tim

moles/time

A

AF

CDefinition of

concentrationfor flow system

FA0

FB0

FC0

FD0

FI0

FA

FB

FC

FD

FI

Entering Leaving

0

0

0

0

00

00

0

0

A

i

A

i

A

i

A

ii y

y

C

C

vC

vC

F

F

)(

)1()1(

0

00

0

Xa

bCC

XCXv

FFC

BAB

AAA

A

CA =f(X)

bB

aAA CkCr

FA =f(X)

Flow Systems

XFadFFC

XFacFFC

XFabFFC

XFF

C

ADDD

ACCC

ABBB

AAA

00

00

00

0

)/(

)/(

)/(

)1(

V

XNadN

V

NC

V

XNacN

V

NC

V

XNabN

V

NC

V

XN

V

NC

ADDD

ACCC

ABBB

AAA

00

00

00

0

)/(

)/(

)/(

)1(

Flow SystemBatch System

If v=vo CA=CAo(1-X)

CB=CAo(B -(b/a)X)

Stoichiometric Table for a Flow System

IAICII

DADADDD

CACACCC

BABAABB

AAAA

ΘFFFΘFI(inert)

Xa

dΘFFX)(F

a

dFΘFD

Xa

cΘFFX)(F

a

cFΘFC

Xa

bΘFFX)(F

a

bFΘFB

X)(FFX)(FFA

000

0000

0000

0000

000 1

(mol/time) (mol/time) (mol/time)

reactor from reactor in reactor toSpecies

rateEffluent Change rate Feed

XFa

b

a

c

a

dFFFTotal ATTT 000 1

Table 3-4 XFFF ATT 00

A0

A0

A0

Volume Change with Reaction

N2 + 3H2 2NH3

- Gas-phase reactions that do not have an equal number of product and reactant moles.

-The combustion chamber of the internal-combustion engine

-The expanding gases within the breech and barrel

Batch Reactor with Variable Volume

Individual concentrations can be determined by expressing the volume V for batch system (or volumetric flow rate v for a flow system) as a function of conversion using the following equation of state.

RTZNPV T

T = temperature, KP = total pressure, atmZ = compressibility factorR = gas constant = 0.08206 dm3·atm/gmol · K

00000 RTNZVP T

This equation is valid at any point in the system at any time. At time t=0,

000

00

T

T

N

N

Z

Z

T

T

P

PVV

Dividing (3-30) by (3-31) and rearranging yields

(3-30)

(3-31)

(3-32)

The total number of moles in the system, NT

XXyXN

N

N

NA

T

A

T

T 111 00

0

0

1a

b

a

c

a

d

XNNN ATT 00

We divide through by NT0

reactedA of mole

moles ofnumber totalin the change

where yA0 is mole fraction of A initially present. If all the species in the generalized reaction are in the gas phase, then

(3-33)

(3-34)

(3-23)

In gas-phase systems that we shall be studying, the temperature and pressure are such that the compressibility factor will not change significantly during the course of the reaction; hence Z0~Z. For a batch system the volume of the gas at any time t is

(3-38)

Eq (3-38) applies only to a variable-volume batch reactor. If the reactor is a rigid steel container of constant volume, then of course V=V0. For a constant-volume container, V=V0, and Eq. (3-38) can be used to calculate the pressure inside the reactor as a function of temperature and conversion.

0

00 )1(

T

TX

P

PVV

Flow Reactor with Variable Volumetric Flow Rate

To derive the concentration of the species in terms of conversion for a variable-volume flow system, we shall use the relationships for the total concentration. The total concentration at any point in the reactor is

(3-39)ZRT

P

v

FC T

T

00

0

0

00 RTZ

P

v

FC T

T

At the entrance to the reactor

(3-40)

Taking Eq (3-40)/Eq(3-39) and assuming Z~Z0,

0

0

00 T

T

P

P

F

Fvv

T

T(3-41)

XFFF ATT 00

From Table 3-4, the total molar flow rate can be written in terms of X

(3-43)

0

000

0

0

0

00

0

0

0

000

11T

T

P

PXyv

T

T

P

PX

F

Fv

T

T

P

P

F

XFFv

AT

A

T

AT

0

00 1

T

T

P

PXv

(3-44)

(3-45)

(3-46)

Recalling yA0=FA0/FT0 and CA0=yA0CT0, then

T

T

P

P

X

XvCC

jjAj

0

0

0

1

Substituting for v using Equation (3-42) and for Fj, we have

Cj = FAo(j+vjX)

vo((1+X)(Po/P)(T/To))Fj v

=

Example 3-5: Determination of Cj=hj(X)for a Gas-Phase Reaction

P3-7BP3-15BP3-16B

Due Date: Next Week

Homework