MSc course Adsorption, Kinetics & Catalysis · 2012. 2. 19. · Adsorption, Kinetics & Catalysis...

1

MSc course

Adsorption, Kinetics & Catalysis

Kinetics

Chapter 2.1-2.6

Prof. Frank de Groot

MSc “Nanomaterials Chemistry and Physics”

Utrecht University

• master / University of Nijmegen / 1987 / Theoretical Chemistry

• PhD / University of Nijmegen / 1991 / Solid State Chemistry

• post-doc / LURE, CNRS Orsay, France / 1993

• post-doc / Groningen (KNAW academy researcher) / 1995

• assistant professor (UD) / Utrecht / 1999 (2001 vidi)

• associate professor (UHD) / Utrecht / 2003 (2006 vici)

• professor / Utrecht / 2009

•

Research

4

Introduction to the course

• Course objectives

– General knowledge on main catalytic processes

– Key concepts in catalysis

• Adsorption

• Kinetics

• Catalysis - mechanisms

• Porosity and surface area – physisorption

• Diffusion

5

Introduction to the course (2)

Steps in diffusion, adsorption and catalysis

Solid catalyst

Pore

6


• Course material

– I. Chorkendorff and J.W. Niemantsverdriet.

“Concepts of Modern Catalysis and Kinetics”, 2nd

Edition, Wiley-VCH (2007)

– Handouts, slides

7


• Course lay-out – see also lecture schedule

– Oil refining, petrochemistry (Ch 9) KdJ

– Synthesis gas (Ch 8) KdJ

– Kinetics (Ch 2) FdG

– Reaction rate theory (Ch 3) FdG

– Diffusion (Ch 5) KdJ

– Surface reactivity (Ch 6) FdG

– Physisorption PdJ

– Examination

• NIOK course “Catalytic Surface Science”, optional

• Rate Equation

• Arrhenius Equation

• Order of a reaction

• Steady State Approximation

• Chain reactions

• Concentration in coupled reactions

Chapter 2 Kinetics (lecture 1)

‘classical kinetics’

Chemical reaction:

Methanol synthesis

• (a, b, p, q) >> stoichiometric coefficients

• (A, B) >> Reactants

• (P, Q) >> Products

• Rate Equation




• Chain reactions


Chapter 2 Kinetics (lecture 1)

‘classical kinetics’

Rate Equation

= change in concentration/

change in time

= d[A]/dt

Chemical reaction




qQpPbBaA

k

k

Chemical reaction




qQpPbBaA

k

k

dt

Bd

bdt

Ad

ar 11

dt

Qd

qdt

Pd

pr 11

Rate equation

rrr

qpbaQPkBAkr

qQpPbBaA

k

k

Equilibrium constant

beq

a

eq

q

eq

p

eq

BA

QP

k

kK

qQpPbBaA

k

k

• Rate Equation




• Chain reactions


Chapter 2 Kinetics

History

• In 1812, the Russian chemist Kirchhof found that when a

water suspension of starch is boiled, no change occurs in the starch. • When a few drops of concentrated sulfuric acid are added to the

same suspension before boiling, the starch breaks down into glucose.

• The acid can be recovered unchanged from the reaction. • Kirchhof concluded that it had played a helping role in the

breakdown of the starch, without itself having undergone any change.

Starch → [H2SO4] → glucose

Arrhenius equation

Arrhenius equation

RTEAAeTk/

)(

Arrhenius equation

RTEAAeTk/

)(

Discovered by:

Jacobus van ‘t Hoff (1884)

Arrhenius equation

Arrhenius equation

RTEAAeTk/

)(

= The fraction of the molecules

present in a gas which have

energies equal to or in excess of

activation energy at a particular

temperature. (Boltzman

distribution; More in chapter 3)

Arrhenius equation

RTEAAeTk/

)(

= A term which includes factors

like the frequency of collisions and

their orientation. It varies slightly

with temperature, although not

much. It is often taken as constant

across small temperature.

(more in chapter 3: reaction rate

theory)

Gas constant R

R= 8.3144 J K-1 mol-1

R = Na * kb

= Avogadro * Boltzmann

= 6·1023 * 1.38 ·10-23

RTEAAeTk/

)(

ºC Rate

0 7.78E-7

25 3.46E-5

45 4.98E-4

55 0.0015

65 0.00487

Arrhenius equation

RTEAAeTk/

)(

Arrhenius equation

RTEAAeTk/

)(

RTEAeATk/

lnln)(ln

RTEATk A /ln)(ln

Arrhenius equation

RTEATk A /ln)(ln

Arrhenius equation

ºC Rate

0 7.78E-7

25 3.46E-5

45 4.98E-4

55 0.0015

65 0.00487

Arrhenius equation

ln k

1/T (K-1)

RTEATk A /ln)(ln

EA= 103 KJ/mol

• Rate Equation




• Chain reactions


Chapter 2 Kinetics

First order reaction

Equation

Time evolution (integrated equation)

Half time

Linearized equation

First order reaction Equation

PRk

Rkrdt

Rd

First order reaction (time evolution)

Rkrdt

Rd

kdtR

Rd

tR

RR

Rddtk

00


tR

RR

Rddtk

00

kteRR 0

kteR

R lnln0


kteRR 0PRk

RRP 0

)1(0

00

kt

kt

eR

eRRP

First order reaction (half time)

kteRR 0

021 RR

021

0 ReR kt

21kte


21kte

21lnln kte

2ln kt

ktt 2ln

21


kteRR 0

S-1

First order reaction (linearize)

kteRR 0

kteR

R 0

kteR

R lnln0


kteR

R lnln0

ktR

R

0

ln

kt

R

R0ln


kt

R

R0ln


Equation

Time evolution (integrated equation)

Half time

Linearized equation


(exercise)

The decomposition reaction SO2Cl2(g) --->

SO2(g) + Cl2(g) is a first order reaction with

rate constant k=2.2 x 10-5 sec-1 at 320C. What

percent of SO2Cl2 is decomposed at 320C after

90 minutes?

Zeroth order reaction

PRk

Excess R

kr

dt

Rd

Zeroth order reaction (time evolution)

tR

R

dtkRd00

ktRR 0

Zeroth order reaction (half time)

021 RR

ktRR 0

021

0 RktR

021 Rkt

k

Rt

20

21


ktRR 0

Reaction order (units)

Order one, the rate coefficient has units of s-1

Order zero, the rate coefficient has units of mol·L-1·s-1

Reaction order (units)

Order one, the rate coefficient has units of s-1

Order zero, the rate coefficient has units of mol·L-1·s-1

Order two, the rate coefficient has units of L·mol-1·s-1

Order n, the rate coefficient has units of mol1-n·Ln-1·s-1

Second order reaction

PRk2

2Rkrdt

Rd

kdt

R

Rd2

Second order reaction (time evolution)

tR

R R

Rddtk

00

2

kt

RR

0

11

Second order reaction (time evolution)

kt

RR

0

11

kt

RR

0

11

kt

R

R

0

1

1

Second order reaction (half time)

021 RR

kt

RR

00

12

kt

R

0

1

kRt

021

1


ktRR 0

kt

R

R

0

1

1

Second order reaction (half time)

Second order reaction (linearize)

kt

RR

0

11

The following data were obtained on the rate of hydrolysis of 17 %

sucrose in 0.99 mol L-1 HCl aqueous solution at 35 C.

t / min 9.82 59.60 93.18 142.9 294.8 589.4

Sucrose remaining, % 96.5 80.3 71.0 59.1 32.8 11.1

What is the order of the reaction with respect to sucrose, and what is the

value of the rate constant k?

Exercise

Zero-Order First-Order Second-Order

Rate Law

Integrated

Rate Law

Linear Plot

to determine k

Half-life

Rkrdt

Rd 2Rkr

dt

Rd

kr

dt

Rd

kt

RR

0

11 ktRR 0 kteRR 0

kt

R

R0ln

kRt

021

1

k

Rt

20

21 k

tt 2ln

21

• Rate Equation




• Chain reactions


Chapter 2 Kinetics

Steady State Approximation

PIRkk 21

PIRk

k

k

2

1

1

PIRkk21


PIRkk21

IkRkdt

Rd 11

IkIkRkdt

Id 211

Ikdt

Pd 2


PIRkk21

0

dt

Id


0211 IkIkRk

RkIkk 121 )(

)( 21

1

kk

RkI

0

dt

Id


0

dt

Id

)( 21

122

kk

RkkIk

dt

Pd

)( 21

1

kk

RkI


RKk

k

Rkkkkdt

Pd12

1

12

)(12

Rk

k

Rkkkkdt

Pd

1

2

12

)(12

PIRkk21

)( 21

122

kk

RkkIk

dt

Pd

Example:

Example:

= 0

Example:

Alternatives to the Steady State Approximation

• Coupled reactions: k-1=0

• Pre-equilibrium solution: [I]=k+1[R]

• Exact solution

PIRkk21

Coupled reactions (concentrations)

PIRkk 21

Rkdt

Rd1

tkeRR 1

0


PIRkk 21

IkRkdt

Id21

)( 21

12

10

tktkee

kk

kRI


PIRkk 21

IRRP 0

)()(1 21

12

1

12

20

tktke

kk

ke

kk

kRP

Figure 2.5

0,0

0,2

0,4

0,6

0,8

1,0

0,0

0,2

0,4

0,6

0,8

1,0

0,0

0,2

0,4

0,6

0,8

1,0

0 1 2 3 4 5 6 7 8 9 10

k1t

k2 = 0.2 k1

k2 = k1

k2 = 10 k1

[R]

[R]

[R]

[P]

[P]

[P]

[I]

[I]

[I]

concentr

ation


I

P R

I

P

R

Alternatives to the Steady State Approximation

• Coupled reactions: k-1=0

• Pre-equilibrium solution: [I]=k+1[R]

• Exact solution

PIRkk21

http://www-jmg.ch.cam.ac.uk/tools/magnus/kinetic.html




• Rate Equation





• Chain reactions

Chapter 2 Kinetics

Chain reaction

OOk

212

NNONOk

22

ONOONk

3

2

242 OO

k

}2{ 25 NN

k

NOON effk222

Chain reaction

2322 ONkNOkdt

NOd

2322 ONkNOkdt

Nd

2421

2322

22 OkOk

ONkNOkdt

Od

Chain reaction: steady state

0

dt

Nd

0

dt

Od


0

dt

Od

0222

421

2322

OkOk

ONkNOk

0222

421 OkOk


0

dt

Od

21

2

4 OkOk

24

1 OOk

k


02322 ONkNOkdt

Nd

0232224

1 ONkNOkk

k

232224

1 ONkNOkk

k


2

2

2

4

1

3

2 NO

ON

k

k

k

k

2

2

4

1

3

2

O

NN

k

k

k

k


2

2

4

1

3

2

O

NN

k

k

k

k

2322 ONkNOkdt

NOd

24

1 OOk

k


2322 ONkNOkdt

NOd

2

2

23

222

4

1

3

2

4

1

OO

Nk

NOk

k

k

k

k

k

k

dt

NOd


2224

12 NOkk

k

dt

NOd

The decomposition of N2O5 → NO2 + NO3 is postulated to take place via two elementary steps:

a. N2O5 + N2O5 N2O5* + N2O5

b. N2O5* → NO2 + NO3

1. Calculate the rate equation, using the steady-state

approximation

2. Assume that reaction b is much faster than the back-

reaction of reaction a. What is the reaction order?

3. Assume that reaction b is much slower than the

back-reaction of reaction a. Reaction order?

Exercise

a. N2O5 + N2O5 N2O5* + N2O5

b. N2O5* → NO2 + NO3

Exercise

*

522

*52521

2521

*52

ONk

ONONk

ONkdt

ONd

Exercise

*522

*52521

2521 ONkONONkONk

)( 2521

2521*

52

kONk

ONkON

Exercise

*522

3 ONkrdt

NOd

)( 2521

25212

kONk

ONkkr

Exercise

)( 2521

25212

kONk

ONkkr

5212

521

25212

)(12

ONKkONk

ONkkr

kk

2521

2

25212

)(12

ONkk

ONkkr

kk

Lindemann-Hinshelwood mechanism

Exercises

Exercises (page 417 and further)

2.5 and 2.6

Questions via email to [email protected]

or room N207 (went)

Next time the exercises will be discussed

Exercises

MSc course Adsorption, Kinetics & Catalysis · 2012. 2. 19. · Adsorption, Kinetics & Catalysis...

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Transcript of MSc course Adsorption, Kinetics & Catalysis · 2012. 2. 19. · Adsorption, Kinetics & Catalysis...