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1

Chemical Reaction Engineering

Chapter 3, Part 1:

Rate Laws

Part I

• Write the relationship between the relative rates of reaction.

• Write a rate law and define reaction order and activation

energy.

Part II

• Set up a stoichiometric table for both batch and flow

systems and express concentration as a function or

conversion.

• Write -rA solely as a function of conversion given the rate

law and then entering concentration.

• Calculate the equilibrium conversion for both gas and liquid

phase reactions.

Rate Law: Objectives

2

Units of k are dependent on the reaction order

The overall order of reaction from units of k

units of k = units of (rA ) /units of conc. terms

Zero-order reaction

rA = k k in mol/m3·s or mol/L·s

First order reaction

rA = kCA k in s-1

Second-order reaction

rA = kCA2 k in m3 /mol·s or L/ mol·s

RECALLING..….

Reaction Rates for Reversible Reactions

• The reaction rate expression should be thermodynamically

consistent.

• That is, at chemical equilibrium (when net rate of reaction is

ZERO) the reaction rate expression must reduce to a form

that satisfy the equilibrium thermodynamic relationship.

Thermodynamic Equilibrium Relationship

aA + bB cC + dD

b

Be

a

Ae

d

De

c

CeC

CC

CCK

Kc – concentration equilibrium

constant

Unit: (mol/dm3) d+c-b-a

3

Reaction Rates for Reversible

Reactions

Reaction: aA + bB cC + dD

krev

kfor

Net rate of formation of A = (rA reverse) + (rA forward) = 0

Rate of consumption of A by forward reaction (-rA_forward)

= kfor CAa CB

b

Rate of formation of A by reverse reaction (rA_reverse)

= krev CCc CD

d

Reaction Rates for Reversible

Reactions

Net rate of formation of A (rA)net = (rA_reverse) + (rA_forward)

(rA)net = krev CCc CD

d - kfor CAa CB

b

We know, at equilibrium net rate = 0

krev (CCc)eq (CD

d)eq - kfor (CAa)eq (CB

b)eq = 0

b

B

a

A

d

D

c

CC

rev

for

CC

CCK

k

k

][)(C

d

D

c

Cb

B

a

AfornetAK

CCCCkr

Same as

thermodynamic

equilibrium

relationship

4

Rate Law: Relative rates of reaction Consider the general equation

The relative rate of reaction of the various species involved in the

reaction can be obtained from the ratio of the stoichiometric

coefficients.

For every mole of A consumed c/a moles of C appears.

Rate of formation of C= (c/a) rate of disappearance of A

The relationship can be expressed directly

from the stoichiometry of the reaction d

r

c

r

b

r

a

r DCBA

Rate Law: Reaction Order

• A rate law describes the behavior of a reaction.

• The rate of a reaction is a function of

temperature (through the rate constant) and

concentration.

• Power Law Model

= n

5

• First Order Reactions

(1) Homogeneous irreversible elementary gas phase reaction

(2) Homogeneous reversible elementary reaction

Examples of Rate Laws

CiCnCn

44

KC-Ckr44

iC nC

360T

360T790631.1expk

333T

333T830.33.03expKC

62HCA

24262

kCr

HHCHC

T

1

1000

1

mol

82kcal

1e0.072sk

Second Order Reactions

(1) Homogeneous irreversible non-elementary

reaction

This is first order in ONCB, first order in

ammonia and overall second order.


3NHONCBA CkCr kmol.min

m0.0017k

3

mol

cal11273E

At 188˚C

6


2323 NCNHBrCHNHCHCNBr

23NHCHCNBrA CkCr s.mol

2.2dmk

3

• Second Order Reactions

(2) Homogeneous irreversible elementary reaction

This reaction is first order in CNBr, first order in

CH3NH2 and overall second order.


(3) Heterogeneous catalytic reaction: The following

reaction takes place over a solid catalyst:

CCBB

PPBCC

PKPK

KPPPkr

1

]/[

7

Part II

• Stoichiometry

Stoichiometry

• If the rate law depends on more than one species we

must relate the concentrations of different species to each

other

• This relationship is most easily established with the aid

of a stoichiometric table

• The stoichiometric table presents the stoichiometric

relationship between reacting molecules for a single

reaction

8

Reaction Stoichiometry

The relative rate of reaction of the various species involved in the

reaction can be obtained from the ratio of the stoichiometric

coefficients.

For every mole of A consumed c/a moles of C appears.

Rate of formation of C= (c/a) × rate of disappearance of A

The relationship can be expressed directly from the stoichiometry

of the reaction

d

r

c

r

b

r

a

r DCBA

Batch Stoichiometric Table

t=0

NA0

NB0

NC0

ND0

NI0

t=t

NA

NB

NC

ND

NI

)(a

b=

reactedA of moles•reactedA of moles

reacted B of moles=

reacted B of moles

X)-(1N=XN-N=N

A of moles initial : N

A isn calculatio of Basis

0

A0A0A0A

A0

XN A

9

Batch Stoichiometric Table

Species Symbols Initial Change Remaining

A A NA0 -NA0X NA=NA0(1–X)

B B NB0=NA0ΘB -(b/a)NA0X NB=NA0(ΘB –(b/a)X)

C C NC0=NA0ΘC (c/a)NA0X NC=NA0(ΘC+(c/a)X)

D D ND0=NA0ΘD (d/a)NA0X ND=NA0(ΘD+(d/a)X)

Inert I NI=NA0ΘI NI = NA0ΘI

NT0 δNA0X

NT=NT0+δNA0X

1 ; 0

0

0

0

0

0 a

b

a

c

a

d

y

y

C

C

N

Ν

A

i

A

i

A

ii

Concentration: Batch Systems

Note: if the reaction occurs in the liquid phase or if a gas phase

reaction occurs in a rigid (e.g., steel) batch reactor Then

V

NC A

A

f(X)–r

Xa

bXCkr

CCkrIf

Xa

bCX

a

b

V

N

V

NC

XCV

XN

V

NC

VV

A

baaa

baaa

BAB

AB

B

A

AA

A

= have weAnd

)-Θ()-1(=-Then

=-

)-Θ(=)-Θ)((==

)-1(=)-1(

==

=

23

0

2

0

0

0

0

0

0

0

Constant Volume Batch:

10

Example

;=Θ

;=Θ

;=Θ

)-1(=

3

1+C→B

3

1+A

)(3

1+→O)(

3

1+NaOH

0

0

0

0

0

0

0

353351735333517

A

D

D

A

C

C

A

B

B

AA

C

C

C

C

C

C

XCC

D

OHHCCOONaHCHCCOOHC


NaOH A NA0 -NA0X NA=NA0(1–X)

C17H35COO)3-

C3H5

B NB0=NA0ΘB -(1/3)NA0X NB=NA0(ΘB –(1/3)X)

C17H35COONa C NC0=NA0ΘC NA0X NC=NA0(ΘC+X)

C3H5(OH)3 D ND0=NA0ΘD (1/3)NA0X ND=NA0(ΘD+(1/3)X)

Water (Inert) I NI=NA0ΘI NI = NA0ΘI

NT0 NT=NT0+δNA0X

11

Flow System Stoichiometric Table

1 ; 0

0

0

0

00

00

0

0 a

b

a

c

a

d

y

y

C

C

vC

vC

F

F

A

i

A

i

A

i

A

ii

Flow System Stoichiometric Table


A A FA0 -FA0X FA=FA0(1–X)

B B FB0=FA0ΘB -(b/a)FA0X FB=FA0(ΘB –(b/a)X)

C C FC0=FA0ΘC (c/a)FA0X FC=FA0(ΘC+(c/a)X)

D D FD0=FA0ΘD (d/a)FA0X FD=FA0(ΘD+(d/a)X)

Inert I FI=FA0ΘI FI = FA0ΘI

FT0 FT=FT0+δFA0X

1 ; 0

0

0

0

00

00

0

0 a

b

a

c

a

d

y

y

C

C

vC

vC

F

F

A

i

A

i

A

i

A

ii

12

Concentration: Flow System

v

FC A

A Liquid Phase Flow System:

)()(

)1()1(

0

0

0

0

0

0

0

Xa

bC

v

Xa

bF

v

FC

XCv

XF

v

FC

vv

BA

BAB

B

AAA

A

If the rate of reaction were:

then we would have:

Xa

bXkCr

CkCr

BAA

BAA

)1(2

0

This gives us

-rA = f(X)

Concentration: Gas Flow System

0

FT

FT0

P0

P

T

T0

FT

FT0

FT0 FA 0X

FT0

1 yA 0 X 1X

FT C T and FT 0 C T 00 C T 0

CT

0

C T P

RT and C T 0

P0

RT0

For gaseous reactants and assuming ideal gases,

therefore from ideal gas law we may get

From

Stoichiometric

Table

Substitute in v

Where:

13

Concentration: Gas Flow System

Combining the compressibility factor equation of state with

Z = Z0

with and

For example if the gas phase reaction has the rate law

then

with

BAA CkCr 2

k

14

Example

Production of Nitric acid

Nitric acid is made commercially from nitric oxide.

Nitric oxide is produced by the gas-phase

oxidation of ammonia.

4NH3 + 5O2 4NO + 6H2O

The feed consists of 15 mol% ammonia in air at 8.2

atm and 227°C.

15

(a) What is the total entering concentration?

Assume ideal gas behavior.

(b) What is the entering concentration of

ammonia?

16

(c) Set up a stoichiometric table with ammonia as your basis

of calculation.

Express Ci for all species as functions of conversion

for a constant-volume batch reactor.

Express PT as a function of X.

Express Pi and Ci for all species as functions of

conversion for a flow reactor.

18

(d) Write the combined mole balance and rate law

solely in terms of the molar flow rates and rate

law parameters for C1 and C2 above. Assume

the reaction is first order in both reactants

20

Summary

1. Rate Laws

– -rA=k f(Ci)

– 1st order A B or 1st order

– 2nd order A+B ==> C

– Rate laws are found by experiment

2. Stoichiometry

– Liquid:

– Gas:

-rA=kCA

rA k C A C B

KC

-rA=kACACB

C A CA0 1 X

C A CA0 1 X

1 X

P

P0

T0

T

C A CT 0

FA

FT

P

P0

T0

T

At the start of the chapter we saw we needed -rA=f(X). This

result is achieved in two steps.

A < === > B

25

General Mole Balance

Batch Reactor Mole Balance

FA0 FA rA dV dNA

dt

FA0 FA 0

dNA

dt rAV

Batch

VrdVr AA Well Mixed

26

Batch Reactor Mole Balance

dt dNA

rAVIntegrating

Time necessary to reduce number of moles of A from NA0 to NA.

when t = 0 NA=NA0

t = t NA=NA

A

A

N

N A

A

Vr

dNt

0

CSTR Mole Balance

FA0 FA rA dV dNA

dt

dNA

dt 0Steady State

CSTR

0 0 VrFF AAA

V FA0 FA

rA

CSTR volum necessary to reduce the molar flow rate

from FA0 to FA.

27

Plug Flow Reactor

0 0 dVrFF AAA

0dt

dN ASteady State

dt

dNdVrFF A

AAA 0

PFR

Plug Flow Reactor Mole

Balance

28

dFA

dV rA

0 dFA

dV rA

Differientiate with respect to V

A

A

F

F A

A

r

dFV

0

The integral form is:

This is the volume necessary to reduce the entering molar flow rate (mol/s)

from FA0 to the exit molar flow rate of FA.

Reactor Mole Balance Summary

Reactor

Differential Algebraic Integral

V FA0 FA

rA

CSTR

Vrdt

dNA

A

0

A

A

N

N A

A

Vr

dNtBatch

NA

t

dFA

dV rA

A

A

F

F A

A

dr

dFV

0

PFR

FA

V

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