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1 - 06/11/22 Department of Chemical Engineerin Lecture 2 Kjemisk reaksjonsteknikk Review of Lecture 1 and 2 (chapter 1 and 2) Rate Laws Reaction Orders Arrhenius Equation Activation Energy Effect of Temperature

Lecture 2 Kjemisk reaksjonsteknikk

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Lecture 2 Kjemisk reaksjonsteknikk. Review of Lecture 1 and 2 (chapter 1 and 2) Rate Laws Reaction Orders Arrhenius Equation Activation Energy Effect of Temperature. General Mole Balance. General Mole Balance on System Volume V. System Volume, V. F A0. F A. G A. 2. - PowerPoint PPT Presentation

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Page 1: Lecture 2 Kjemisk reaksjonsteknikk

1 - 04/21/23

Departm

ent of Chem

ical E

ngineering

Lecture 2

Kjemisk reaksjonsteknikk

Review of Lecture 1 and 2 (chapter 1 and 2) Rate Laws Reaction Orders Arrhenius Equation Activation Energy Effect of Temperature

Page 2: Lecture 2 Kjemisk reaksjonsteknikk

2 - 04/21/23

Departm

ent of Chem

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ngineering

General Mole Balance

General Mole Balance on System Volume V

In Out Generation Accumulation

FA 0 FA rA dV dNA

dt

FA

0

FAGA

System Volume, V

2

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Reactor Mole Balance Summary

Reactor Differential Algebraic Integral

The GMBE applied to the four major reactor types (and the general reaction AB)

V FA 0 FA

rA

CSTR

Vrdt

dNA

A 0

A

A

N

N A

A

Vr

dNtBatch

NA

t

dFA

dVrA

A

A

F

F A

A

dr

dFV

0

PFRFA

V

dFA

dW r A

A

A

F

F A

A

r

dFW

0

PBRFA

W3

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Reactor Mole Balances in terms of conversion

Reactor Differential Algebraic Integral

A

0A

r

XFV

CSTR

A0A rdV

dXF

X

0 A0A r

dXFVPFR

Vrdt

dXN A0A

Vr

dXNt

X

0 A0A

Batch

X

t

A0A rdW

dXF

X

0 A0A r

dXFWPBR

X

W4

FA=FA0-FA0X=FA0(1-X)

X

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Reactor Mole Balances in terms of residence time and conversion

X

t

5

FA=CAΦV , CA=CA0(1-X), τ=V/ ΦV

X

Batch reactor dt

dxC

dt

dCr A

AA 0

CSTR

xCr A

A0

PFR d

dxC

FVd

dxr A

AA 0

0 )/(

X

τ

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Two types of problems Design problem: Design a reactor to achieve certain conversion

Operating problem: in a existing reactor to find conversion or the residence time to reach the certain conversion

A

x

AA r

dxC

FVd

dx

0

00 )/(

PFR

Levenspiel Plots

A

A

r

xC

0

CSTR

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Reactors in Series

2,

1202

)(

xA

A

r

xxFV

1,

101

xA

A

r

xFV

1

0

01

x

AA r

dxFV

1,

101

xA

A

r

xFV

2

1

02

x

x AA r

dxFV

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PFR vs CSTR in series

…..

A

A

r

F

0

x

Is the PFR always better than the CSTR in terms of reactor size to achieve a identical conversion?

i

iAA V

xFr 0

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Reactors in Series

9

reactorfirst tofedA of moles

ipoint toup reactedA of molesX i

Only valid if there are no side streams

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Reactors in Series

10

Exothermic reaction in an adiabatic reactor

How can we minimize the reaction size?

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Basic Definitions

Homogeneous reactions involve only one phase Heterogeneous reactions involve more than one phase, and

reactions occur at interfaces of two phases Irreversible reactions occur at only one direction Reversible reactions occur at both directions, depending one the

approach to equilibrium Molecularity of the reaction is the number of the atoms, ions or

molecules involved in a reaction step. Unimolecular, bimolecular and termolecular refer to reactions involving one, two and three atoms or molecules in one reaction step, respectively

Elementary reaction involves only one bond breaking or formation Non-elementary reaction could involve multi elementary reaction

steps

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Kinetics

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BAA CkCr

βα

β

α

OrderRection Overall

Bin order

Ain order

Kinetics - Power Law Model

13

C3BA2 A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written.e.g. If the above reaction follows an elementary rate law

2nd order in A, 1st order in B, overall third order

B2AAA CCkr

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"Everyone has Problems -

but Chemists have Solutions"

Chemical Engineers have Simple Solutions !!!

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Example : Ammonia decomposition

2NH3=3H2+N2 11 kJ/mol

The kinetic study was performed in a fixed bed reactor (6 mm diameter) on Fe/Al2O3 catalysts at atmospheric pressure and total flow of 100 ml/min with Ar as the balance. 100 mg catalysts mixed with 1 g SiC were loaded in the reactor.

FNH3.s (ml/min) 20 40 80

FAr,s (ml/min) 80 60 20

XNH3 0.050 0.051 0.050

1) Can we determine the reactor order? (n=0,1,2 ? ) 2) How can we reduce the conversion from 5.0 % to 2.5 %

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dDcCbBaA

d

r

c

r

b

r

a

r DCBA

Da

dC

a

cB

a

bA

Relative Rates of Reaction

16

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3

r

1

r

2

r CBA

Relative Rates of Reaction

C3BA2

sdm

mol10r

3A

sdm

mol5

2

rr

3A

B

sdm

mol15r

2

3r

3AC

17

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If2AA kCr

›Second Order in A›Zero Order in B›Overall Second Order

2A+B3C

BAAA CCkr 2

BABB CCkr 2

2

1

BACC CCkr 2

2

3

18

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Reversible Reaction

Elementary

e

3C2

BAA

AA

3C2

BAA

3CA

2BAAA

K

CCCk

kk

CCCk

CkCCkr

19

C3BA2

This equation is thermodynamically consistent.

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Reversible Reaction

A+2B3C

kA

k-AReaction is: First Order in A

Second Order in BOverall third Order

sm

molesrA 3

3m

molesCA

smole

m

mmolemmole

smmole

CC

rk

BA

A2

6

233

3

2

20

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Arrhenius Equation

k Ae E RT

k is the specific reaction rate (constant) and is given by the Arrhenius Equation.

where:

T k A

T 0 k 0

A 1013

k

T22

Svante August Arrhenius was a Swedish scientist, received the Nobel Prize for Chemistry in 1903

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Arrhenius Equation

where:E = Activation energy (cal/mol)R = Gas constant (cal/mol*K)T = Temperature (K)A = Frequency factor (same units as rate constant k)(units of A, and k, depend on overall reaction order)

23

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Reaction Coordinate

The activation energy can be thought of as a barrier to the reaction. One way to view the barrier to a reaction is through the reaction coordinates. These coordinates denote the energy of the system as a function of progress along the reaction path. For the reaction:

A BC A ::: B ::: C AB CThe reaction coordinate is

24

Transition state theory

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Why is there an Activation Energy?

We see that for the reaction to occur, the reactants must overcome an energy barrier or activation energy EA. The energy to overcome their barrier comes from the transfer to the kinetic energy from molecular collisions and internal energy (e.g. Vibrational Energy).1.The molecules need energy to disort or

stretch their bonds in order to break them and thus form new bonds

2.As the reacting molecules come close together they must overcome both stearic and electron repulsion forces in order to react.25

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Collision probability

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One such distribution of energies is in the following figure:

f(E,T)dE=fraction of molecules with energies between E+dE

29

Page 30: Lecture 2 Kjemisk reaksjonsteknikk

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Rate expression – Gas phase reaction

A+B = 2C

Page 31: Lecture 2 Kjemisk reaksjonsteknikk

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Rate expression – Catalysed reaction