Catalytic Reaction Engineering - Aalto · Derive reaction rate law when propene (PE) adsorbs...

CatalyticReaction Engineering

Yongdan Li

Nov-Dec, 2018

Professor of Industrial ChemistryDepartment of Chemical and Metallurgical EngineeringSchool of Chemical TechnologyAalto UniversityEmail: [email protected] 1, E404

Reaction Mechanism

and Rate Equations

of Heterogeneous

Catalytic Reactions

2

1. Bulk fluid → External surface

2. External surface → Internal surface

3. Surface adsorption

4. Surface reaction

5. Surface desorption

6. Internal surface → External surface

7. External surface → Bulk fluid

●The overall rate of reaction is limited by the rate of the slowest

step in the sequence.

●The slowest step is called rate- determining step (RDS).

Heterogeneous catalysis – reaction mechanisms

3

A

A: atom, molecule, or some other

atomic combination.

S: active site

Rate of attachment = kA PA CV

A+S A·SkA

k-A

Rate of detachment = k-A CA·S

Pi: partial pressure of species i in the

gas phase

Ci·S: surface concentration of sites

occupied by species i

CV: surface concentration of vacant

sites

KA: Rate constant of the attachment

processes

K-A: Rate constant of the detachment

processes

Ct: the total concentration of sites

Ct = CA·S + CV

Adsorption

4

Two types of adsorption:

Dissociative adsorptionMolecular adsorption

Adsorption

5

1. Molecular adsorption

CO+S CO·SkA

k-A

𝒓𝑨𝑫 = 𝒌𝑨 𝑷𝑪𝑶𝑪𝑽 −𝑪𝑪𝑶∙𝑺𝑲𝑨

Adsorption

𝑪𝑪𝑶∙𝑺 =𝑲𝑨𝑷𝑪𝑶𝑪𝒕𝟏 + 𝑲𝑨𝑷𝑪𝑶

2. Dissociative adsorption

CO + 2S C·S + O·SkA

k-A

𝑪𝑶∙𝑺 =𝑲𝑨𝑷𝑪𝑶

Τ𝟏 𝟐𝑪𝒕

𝟏 + 𝟐 𝑲𝑨𝑷𝑪𝑶Τ𝟏 𝟐

𝒓𝑨𝑫 = 𝒌𝑨 𝑷𝑪𝑶𝑪𝑽𝟐 −

𝑪𝑪∙𝑺𝑪𝑶∙𝑺𝑲𝑨

6

After adsorption, surface reaction can take place via

– Single site mechanism (Langmuir-Hinshelwood)

– Dual site mechanism (Langmuir-Hinshelwood)

– Adsorbed molecule can react with gas-phase component (Eley-

Rideal)

Surface Reaction

7

Single site mechanism

In single site mechanism only one active surface site is involved

A ↔ BFor example, isomerization

Surface Reaction ● Langmuir-Hinshelwood

𝐴 + 𝑆 𝐴 ∙ S𝑘𝐴

𝑘−𝐴

𝐵 ∙ 𝑆 𝐵 + 𝑆𝑘𝐵

𝑘−𝐵

A

B𝐴 ∙ 𝑆 𝐵 ∙ 𝑆

𝑘+𝑆

𝑘−𝑆

8

Rate of surface reaction:

𝒓𝑺 = 𝒌+𝒔𝑪𝑨∙𝑺 − 𝒌−𝒔𝑪𝑩∙𝑺 𝐶𝐴∙𝑆 =𝐾𝐴𝑃𝐴𝑪𝒕

1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵

𝐶𝐵∙𝑆 =𝐾𝐵𝑃𝐵𝑪𝒕

1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵= 𝒌+𝒔 𝑪𝑨∙𝑺 −

𝑪𝑩∙𝑺𝑲𝒔

Where 𝑲𝒔 is the surface-reaction equilibrium constant

𝐾𝑆 = Τ𝑘+𝑆 𝑘−𝑆


9

Dual site mechanism

• In dual site mechanism two active surface site are involved

– Adsorbed component A reacts with unoccupied site:

A∙S + S ↔ B∙S + S

– Adsorbed component A reacts with another adsorbed

component B:

A∙S + B∙S ↔ C∙S + D∙S

– Also two different sites can be involved:

A∙S + B∙S’ ↔ C∙S + D∙S’


10

A

B

1. Adsorbed component A reacts with unoccupied site as the

rate determining step:


𝑘−𝐴


𝑘−𝐵

𝑨 ∙ 𝑺 + 𝑺 𝑩 ∙ 𝑺 + 𝑺𝒌+𝑺

𝒌−𝑺



𝒓𝑺 = 𝒌+𝒔𝑪𝑨∙𝑺𝑪𝑽 − 𝒌−𝒔𝑪𝑩∙𝑺𝑪𝑽 = 𝒌+𝒔 𝑪𝑨∙𝑺𝑪𝑽 −𝑪𝑩∙𝑺𝑪𝑽𝑲𝒔

𝐶𝐴∙𝑆 =𝐾𝐴𝑃𝐴𝑪𝒕

1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵𝐶𝐵∙𝑆 =

𝐾𝐵𝑃𝐵𝑪𝒕1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵

11


2. Adsorbed component A reacts with another adsorbed

component B:


𝑘−𝐴

𝐶 ∙ 𝑆 𝐶 + 𝑆𝑘𝐶

𝑘−𝐶

𝑨 ∙ 𝑺 + 𝑩 ∙ 𝑺 𝑪 ∙ 𝑺 + 𝑫 ∙ 𝑺𝒌+𝑺

𝒌−𝑺

𝐵 + 𝑆 𝐵 ∙ S𝑘𝐵

𝑘−𝐵

𝐷 ∙ 𝑆 𝐷 + 𝑆𝑘𝐷

𝑘−𝐷

A B

C D

12



1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵 + 𝐾𝑐𝑃𝑐 + 𝐾𝑑𝑃𝑑

𝒓𝑺 = 𝒌+𝒔𝑪𝑨∙𝑺𝑪𝑩∙𝑺 − 𝒌−𝒔𝑪𝑪∙𝑺𝑪𝑫∙𝑺 = 𝒌+𝒔 𝑪𝑨∙𝑺𝑪𝑩∙𝑺 −𝑪𝑪∙𝑺𝑪𝑫∙𝑺

𝑲𝒔


𝐶𝐵∙𝑆 =𝐾𝐵𝑃𝐵𝑪𝒕


𝐶𝐶∙𝑆 =𝐾𝐶𝑃𝐶𝑪𝒕


𝐶𝐷∙𝑆 =𝐾𝐷𝑃𝐷𝑪𝒕


13

3. Two species adsorbed on different types of sites S and S’:


𝑘−𝐴

𝐶 ∙ 𝑆 𝐶 + 𝑆𝑘𝐶

𝑘−𝐶

𝐴 ∙ 𝑆 + 𝐵 ∙ 𝑆′ 𝐶 ∙ 𝑆 + 𝐷 ∙ 𝑆′𝑘+𝑆

𝑘−𝑆

𝐵 + 𝑆′ 𝐵 ∙ S′𝑘𝐵

𝑘−𝐵

𝐷 ∙ 𝑆′ 𝐷 + 𝑆′𝑘𝐷

𝑘−𝐷


A B

C D

14


𝒓𝑺 = 𝒌+𝒔𝑪𝑨∙𝑺𝑪𝑩∙𝑺′ − 𝒌−𝒔𝑪𝑪∙𝑺𝑪𝑫∙𝑺′ = 𝒌+𝒔 𝑪𝑨∙𝑺𝑪𝑩∙𝑺′ −𝑪𝑪∙𝑺𝑪𝑫∙𝑺′

𝑲𝒔



1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐶𝑃𝐶𝐶𝐵∙𝑆′ =

𝐾𝐵𝑃𝐵𝑪𝒕1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐷𝑃𝐷

𝐶𝐶∙𝑆 =𝐾𝐶𝑃𝐶𝑪𝒕

1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐶𝑃𝐶𝐶𝐷∙𝑆′ =

𝐾𝐷𝑃𝐷𝑪𝒕1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐷𝑃𝐷

15

Surface Reaction ● Eley-Rideal

Adsorbed molecule react with gas-phase component


𝑘−𝐴

𝐶 ∙ 𝑆 𝐶 + 𝑆𝑘𝐵

𝑘−𝐵

𝑨 ∙ 𝑺 + 𝑩 𝑪 ∙ 𝑺𝒌+𝑺

𝒌−𝑺

A

C

B

𝒓𝑺 = 𝒌+𝒔𝑪𝑨∙𝑺𝑷𝑩 − 𝒌−𝒔𝑪𝑪∙𝑺 = 𝒌+𝒔 𝑪𝑨∙𝑺𝑷𝑩 −𝑪𝑪∙𝑺𝑲𝒔



1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐶𝑃𝐶𝐶𝐶∙𝑆 =

𝐾𝐶𝑃𝐶𝑪𝒕1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐶𝑃𝐶

16

Desorption

C·S C+SkD

k-D

the rate of desorption of C is

𝒓D = 𝒌𝑫𝑪𝑪∙𝑺 − 𝒌−𝑫𝑷𝑪𝑪𝑽 = 𝒌𝑫 𝑪𝑪∙𝑺 −𝑷𝑪𝑪𝑽𝑲𝑫𝑪

C

17

• Assumption: Fast and slow reaction steps

• Slowest step determines overall reaction rate

– Changes in slowest step decides reaction rate

• Fast steps are in equilibrium:

E.g. 𝑟𝐴𝐷 = 𝑟+𝐴𝐷 − 𝑟−𝐴𝐷 = 0

Rate-Determining Step

18

𝐴 + 𝑆 𝐴 ∙ 𝑆𝑘𝐴

𝑘−𝐴

𝐴 ∙ 𝑆 𝐵 ∙ 𝑆 + 𝐶𝑘𝑆

𝑘−𝑆


𝑘−𝐵

𝑨→𝑩 + 𝑪Reaction：

Step：


19

𝒓𝑨𝑫 = 𝒌𝑨𝑷𝑨𝑪𝑽 − 𝒌−𝑨𝑪𝑨∙𝑺 = 𝒌𝑨 𝑷𝑨𝑪𝑽 −𝑪𝑨∙𝑺𝑲𝑨

The rate of adsorption is

The rate of reaction is

𝒓𝑺 = 𝒌𝑺𝑪𝑨∙𝑺 − 𝒌−𝑺𝑷𝑪𝑪𝑩∙𝑺 = 𝒌𝑺 𝑪𝑨∙𝑺 −𝑷𝑪𝑪𝑩∙𝑺𝑲𝑺

The rate of desorption is

𝒓𝑫 = 𝒌𝑩𝑪𝑩∙𝑺 − 𝒌−𝑩𝑷𝑩𝑪𝑽 = 𝒌𝑩 𝑪𝑩∙𝑺 −𝑷𝑩𝑪𝑽𝑲𝑫

𝐾𝐴 = Τ𝑘𝐴 𝑘−𝐴

𝐾𝑆 = Τ𝑘𝑆 𝑘−𝑆

𝐾𝐷 = Τ𝑘𝐵 𝑘−𝐵

The B adsorption equilibrium constant KB is just the reciprocal of the B

desorption constant KD

𝐾𝐵 =1

𝐾𝐷

The rate of desorption can be written as 𝒓𝑫 = 𝒌𝑩 𝑪𝑩∙𝑺 −𝑲𝑩𝑷𝑩𝑪𝑽


20

1. Adsorption is the RDS

−𝒓𝑨′ = 𝒓𝑨𝑫= 𝒌𝑨 𝑷𝑨𝑪𝑽 −

𝑪𝑨∙𝑺𝑲𝑨

For adsorption-limited reaction, 𝒌𝑨 is very small and 𝒌𝑺, 𝒌𝑩 are very large

in comparison. Consequently, the ratios Τ𝒓𝑺 𝒌𝑺, Τ𝒓𝑫 𝒌𝑩 are very small

(approximately zero), whereas the ratio Τ𝒓𝑨𝑫 𝒌𝑨 is relatively large.


21

𝒓𝑺𝒌𝑺

≈ 𝟎𝒓𝑺 = 𝒌𝑺 𝑪𝑨∙𝑺 −𝑷𝑪𝑪𝑩∙𝑺𝑲𝑺

So: 𝑪𝑨∙𝑺 =𝑷𝑪𝑪𝑩∙𝑺𝑲𝑺

Reaction:

𝒓𝑫 = 𝒌𝑩 𝑪𝑩∙𝑺 −𝑲𝑩𝑷𝑩𝑪𝑽𝒓𝑫𝒌𝑩

≈ 𝟎 So: 𝑪𝑩∙𝑺 = 𝑲𝑩𝑷𝑩𝑪𝑽

Desorption:

So: 𝑪𝑨∙𝑺 =𝑲𝑩𝑷𝑩𝑷𝑪

𝑲𝑺𝑪𝑽

Replacing 𝑪𝑨∙𝑺 in the rate equation and then factoring 𝑪𝑽, we obtain:

𝒓𝑨𝑫 = 𝒌𝑨 𝑷𝑨 −𝑲𝑩𝑷𝑩𝑷𝑪𝑲𝑨𝑲𝑺

𝑪𝑽 = 𝒌𝑨 𝑷𝑨 −𝑷𝑩𝑷𝑪𝑲𝑷

𝑪𝑽

Where: 𝐾𝑃 =𝐾𝐴𝐾𝑆𝐾𝐵

Total amount of sites: 𝐶𝑡 = 𝐶𝑉 + 𝐶𝐴∙𝑆 + 𝐶𝐵∙𝑆

Substituting for 𝑪𝑩∙𝑺 and 𝑪𝑨∙𝑺 in the Equation :

𝐶𝑡 = 𝐶𝑉 +𝐾𝐵𝑃𝐵𝑃𝐶𝐾𝑆

𝐶𝑉 + 𝐾𝐵𝑃𝐵𝐶𝑉


22

𝐶𝑉 =𝐶𝑡

1 + Τ𝐾𝐵𝑃𝐵𝑃𝐶 𝐾𝑆 + 𝐾𝐵𝑃𝐵

−𝑟𝐴′ = 𝑟𝐴𝐷= 𝑘𝐴 𝑃𝐴𝐶𝑉 −

𝐶𝐴∙𝑆𝐾𝐴

= 𝑘𝐴 𝑃𝐴 −𝑃𝐵𝑃𝐶𝐾𝑃

𝐶𝑉So:

=𝐶𝑡𝑘𝐴 𝑃𝐴 − Τ𝑃𝐵𝑃𝐶 𝐾𝑃

1 + Τ𝐾𝐵𝑃𝐵𝑃𝐶 𝐾𝑆 + 𝐾𝐵𝑃𝐵


Solving for amount of vacant sites:


23

𝒓𝑫𝒌𝑩

≈ 𝟎

2. Surface Reaction is the RDS

−𝒓𝑨′ = 𝒓𝑺 = 𝒌𝑺 𝑪𝑨∙𝑺 −

𝑷𝑪𝑪𝑩∙𝑺𝑲𝑺

𝒓𝑫 = 𝒌𝑩 𝑪𝑩∙𝑺 −𝑲𝑩𝑷𝑩𝑪𝑽 So: 𝑪𝑩∙𝑺 = 𝑲𝑩𝑷𝑩𝑪𝑽

Desorption:

Adsorption

𝒓𝑨𝑫 = 𝒌𝑨 𝑷𝑨𝑪𝑽 −𝑪𝑨∙𝑺𝑲𝑨

𝒓𝑨𝑫𝒌𝑨

≈ 𝟎 So: 𝑪𝑨∙𝑺 = 𝑲𝑨𝑷𝑨𝑪𝑽


24

Substituting for 𝑪𝑩∙𝑺 and 𝑪𝑨∙𝑺 in the rate equation, we obtain:

𝒓𝑺 = 𝒌𝑺 𝑲𝑨𝑷𝑨 −𝑲𝑩𝑷𝑩𝑷𝑪

𝑲𝑺𝑪𝑽 = 𝒌𝑺𝑲𝑨 𝑷𝑨 −

𝑷𝑩𝑷𝑪𝑲𝑷

𝑪𝑽


𝐶𝑡 = 𝐶𝑣 + 𝐶𝐴∙𝑆 + 𝐶𝐵∙𝑆

Substituting for 𝑪𝑩∙𝑺 and 𝑪𝑨∙𝑺 ,factoring out 𝑪𝒗 :

𝐶𝑉 =𝐶𝑡


Total amount of sites:


25

So:

=𝐶𝑡𝑘𝑆𝐾𝐴 𝑃𝐴 − Τ𝑃𝐵𝑃𝐶 𝐾𝑃


Where: 𝑘 = 𝐶𝑡𝑘𝑆𝐾𝐴

−𝑟𝐴′ = 𝑟𝑆 = 𝑘𝑆𝐾𝐴 𝑃𝐴 −

𝑃𝐵𝑃𝐶𝐾𝑃

𝐶𝑉

=𝑘 𝑃𝐴 − Τ𝑃𝐵𝑃𝐶 𝐾𝑃1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵


26

3. Desorption is the RDS

−𝒓𝑨′ = 𝒓𝑫 = 𝒌𝑩 𝑪𝑩∙𝑺 −𝑲𝑩𝑷𝑩𝑪𝑽

Adsorption

𝒓𝑨𝑫 = 𝒌𝑨 𝑷𝑨𝑪𝑽 −𝑪𝑨∙𝑺𝑲𝑨

𝒓𝑨𝑫𝒌𝑨

≈ 𝟎 So: 𝑪𝑨∙𝑺 = 𝑲𝑨𝑷𝑨𝑪𝑽

𝒓𝑺𝒌𝑺

≈ 𝟎𝒓𝑺 = 𝒌𝑺 𝑪𝑨∙𝑺 −𝑷𝑪𝑪𝑩∙𝑺𝑲𝑺

So: 𝑪𝑩∙𝑺 =𝑲𝑺𝑪𝑨∙𝑺𝑷𝑪

Reaction:


27

Substituting for 𝑪𝑩∙𝑺 and 𝑪𝑨∙𝑺 in the rate equation, we obtain:

𝒓𝑫 = 𝒌𝑩𝑲𝑨𝑲𝑺𝑷𝑨

𝑷𝑪− 𝑲𝑩𝑷𝑩 𝑪𝑽 = 𝒌𝑩𝑲𝑨𝑲𝑺

𝑷𝑨𝑷𝑪

−𝑷𝑩𝑲𝑷

𝑪𝑽


𝐶𝑡 = 𝐶𝑣 + 𝐶𝐴∙𝑆 + 𝐶𝐵∙𝑆

Substituting for 𝑪𝑩∙𝑺 and 𝑪𝑪∙𝑺 ,factoring out 𝑪𝑽 :

𝐶𝑉 =𝐶𝑡

1 + Τ𝐾𝐴𝐾𝑆𝑃𝐴 𝑃𝐶 + 𝐾𝐴𝑃𝐴



28

So:

=𝐶𝑡𝑘𝐵𝐾𝐴𝐾𝑆 Τ𝑃𝐴 𝑃𝐶 − Τ𝑃𝐵 𝐾𝑃1 + 𝐾𝐴𝑃𝐴 + Τ𝐾𝐴𝐾𝑆𝑃𝐴 𝑃𝐶

Where: 𝑘 = 𝐶𝑡𝑘𝐵𝐾𝐴𝐾𝑆

−𝑟𝐴′ = 𝑟𝐷 = 𝑘𝐵𝐾𝐴𝐾𝑆

𝑃𝐴𝑃𝐶

−𝑃𝐵𝐾𝑃

𝐶𝑉

=𝑘 Τ𝑃𝐴 𝑃𝐶 − Τ𝑃𝐵 𝐾𝑃

1 + 𝐾𝐴𝑃𝐴 + Τ𝐾𝐴𝐾𝑆𝑃𝐴 𝑃𝐶


29

−𝑟𝐴′ =

𝑘 𝑃𝐴 − Τ𝑃𝐵𝑃𝐶 𝐾𝑃1 + Τ𝐾𝐵𝑃𝐵𝑃𝐶 𝐾𝑆 + 𝐾𝐵𝑃𝐵

=𝒌𝑷𝑨

𝟏 + Τ𝑲𝑩𝑷𝑩𝑷𝑪 𝑲𝑺 +𝑲𝑩𝑷𝑩

−𝑟𝐴′ =

𝑘 𝑃𝐴 − Τ𝑃𝐵𝑃𝐶 𝐾𝑃1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵

=𝒌𝑷𝑨

𝟏 + 𝑲𝑨𝑷𝑨 +𝑲𝑩𝑷𝑩

−𝑟𝐴′ =

𝑘 Τ𝑃𝐴 𝑃𝐶 − Τ𝑃𝐵 𝐾𝑃1 + 𝐾𝐴𝑃𝐴 + Τ𝐾𝐴𝐾𝑆𝑃𝐴 𝑃𝐶

=𝒌 Τ𝑷𝑨 𝑷𝑪

𝟏 + 𝑲𝑨𝑷𝑨 + Τ𝑲𝑨𝑲𝑺𝑷𝑨 𝑷𝑪

Adsorption is RDS:

Surface reaction is RDS:

Desorption is RDS:

Overall equilibrium constant is the product of the equilibrium

constants of the steps, thus in this case:

For irreversible reaction Kp is large, so 1/Kp is small


30

𝑲𝒑 =𝑲𝑨 ∙ 𝑲𝑺

𝑲𝑩

−𝒓𝑨′ =

𝒌𝑷𝑨𝟏 + Τ𝑲𝑩𝑷𝑩𝑷𝑪 𝑲𝑺 + 𝑲𝑩𝑷𝑩

→ −𝒓𝑨𝟎′ = 𝒌𝑷𝑨𝟎

−𝒓𝑨′ =

𝒌𝑷𝑨𝟏 + 𝑲𝑨𝑷𝑨 +𝑲𝑩𝑷𝑩

→ −𝒓𝑨𝟎′ =

𝒌𝑷𝑨𝟎𝟏 + 𝑲𝑨𝑷𝑨𝟎

−𝒓𝑨′ =

𝒌 Τ𝑷𝑨 𝑷𝑪𝟏 + 𝑲𝑨𝑷𝑨 + Τ𝑲𝑨𝑲𝑺𝑷𝑨 𝑷𝑪

→ −𝒓𝑨𝟎′ = 𝒌′

Adsorption is RDS:


Desorption is RDS:

Initial rate of reaction:


31

-r’ A

0

PA0

-r’ A

0

PA0

-r’ A

0

PA0

Desorption is RDS:

−𝒓𝑨𝟎 = 𝒌′

Adsorption is RDS:

−𝒓𝑨𝟎 = 𝒌𝑷𝑨𝟎−𝒓𝑨𝟎 =

𝒌𝑷𝑨𝟎𝟏 + 𝑲𝑨𝑷𝑨𝟎

−𝒓𝑨𝟎 = 𝒌𝑷𝑨𝟎

−𝒓𝑨𝟎 =𝒌𝑷𝑨𝟎𝑲𝑨𝑷𝑨𝟎

= 𝒌′

At low partial pressures of A: 1 » KAPC0

At high partial pressures of A: 1 « KAPC0


Reaction： 𝑨→𝑩 + 𝑪


32


33

Derivation of rate laws

• Assumptions needs to be made on

1. Adsorption

• Molecular or dissociative

• Same or different active sites (competitive adsorption or not)

2. Mechanism of surface reaction

• Langmuir-Hinshelwood (single or dual site)

• Eley-Rideal

3. Desorption

4. Rate-Determining step

• Reaction rate laws for each step

• Elimination of unknowns

34


Instructions:

1. Make rate law for each step (steps are elementary)

2. Reaction rate constant of RDS is small

→ For others -r/k ≈ 0

→ Solve surface concentrations

3. Make site balance and solve Cv

4. Substitute surface concentrations and Cv into rate law

5. Substitute Kp (if overall reaction is irreversible)

6. Compare the rate law with experimental data

35

𝐶 + 𝑆 𝐶 ∙ S𝑘𝐴

𝑘−𝐴

𝐶 ∙ 𝑆 𝐵 ∙ 𝑆 + 𝑃𝑘𝑆

𝑘−𝑆

𝐵 ∙ 𝑆 𝐵 + 𝑆𝑘𝐷

𝑘−𝐷

𝑪𝟔𝑯𝟓𝑪𝑯(𝑪𝑯𝟑)𝟐→𝑪𝟔𝑯𝟔 + 𝑪𝟑𝑯𝟔Example 1:

Step：

B=benzene

P=propylene

C=cumene


36

−𝑟𝐶′ =

𝐶𝑡𝑘𝐴 𝑃𝐶 − Τ𝑃𝐵𝑃𝑃 𝐾𝑃1 + Τ𝐾𝐵𝑃𝐵𝑃𝑃 𝐾𝑆 + 𝐾𝐵𝑃𝐵

−𝑟𝐶′ =

𝐶𝑡𝑘𝑆𝐾𝐶 𝑃𝐶 − Τ𝑃𝐵𝑃𝑃 𝐾𝑃1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐶𝑃𝐶

−𝑟𝐶′ =

𝐶𝑡𝑘𝐷𝐾𝐶𝐾𝑆 𝑃𝐶 − Τ𝑃𝐵𝑃𝑃 𝐾𝑃𝑃𝑃 + 𝐾𝐶𝑃𝐶𝑃𝑃 + 𝐾𝐶𝐾𝑆𝑃𝐶

Adsorption is RDS:


Desorption is RDS:

Which rate law is correct?

Actual initial rate as a function of partial

pressure of cumene.

The rate law derived by assuming

that the surface reaction is rate-

Determining agrees with the data.


37


The rate law for the case of no inerts adsorbing on the surface is:

−𝑟𝐶′ =

𝐶𝑡𝑘𝑆𝐾𝐶 𝑃𝐶 − Τ𝑃𝐵𝑃𝑃 𝐾𝑃1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐶𝑃𝐶

If we were to have an adsorbing inert in the feed, the inert would not

participate in the reaction but would occupy active sites on the catalyst

surface:

𝐼 + 𝑆 𝐼 ∙ S

𝐶𝑡 = 𝐶𝑣 + 𝐶𝐴∙𝑆 + 𝐶𝐵∙𝑆 + 𝐶𝐼∙𝑆The site balance :

Because the adsorption of the inert is at equilibrium, the concentration of

sites occupied by the inert is

𝑪𝑰∙𝑺 = 𝑲𝑰𝑷𝑰𝑪𝑽

38

Substituting for the inert sites in the site balance, the rate law for surface

reaction control when an adsorbing inert is present is

−𝑟𝐶′ =

𝐶𝑡𝑘𝑆𝐾𝐶 𝑃𝐶 − Τ𝑃𝐵𝑃𝑃 𝐾𝑃1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐶𝑃𝐶 + 𝐾𝐼𝑃𝐼

=𝑘 𝑃𝐶 − Τ𝑃𝐵𝑃𝑃 𝐾𝑃

1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐶𝑃𝐶 + 𝐾𝐼𝑃𝐼


39


Example 2: Propene hydrogenation

Propene (PE) + H2 → Propane (PA)

Derive reaction rate law when propene (PE) adsorbs molecularly, surface

reaction is rate-determining step, propane (PA) is formed on active site

and mechanism is

1. Langmuir-Hinshelwood, hydrogen adsorbs dissociatively

2. Eley-Rideal (hydrogen reacts directly from gas-phase)

3. Langmuir-Hinshelwood, hydrogen adsorbs molecularly

40


PE + S PE ∙ S𝑘𝑃𝐸

𝑘−𝑃𝐸

PA ∙ S PA + S𝑘𝑃𝐴

𝑘−𝑃𝐴

H2 + 2S 2H ∙ S𝑘𝐻2

𝑘−𝐻2

PE ∙ S + 2H ∙ S PA ∙ S + 2S𝑘𝑆


41

Adsorption:

Surface reaction:

Desorption

𝒓𝑨𝑷𝑬 = 𝒌𝑷𝑬𝑷𝑷𝑬𝑪𝑽 − 𝒌−𝑷𝑬𝑪𝑷𝑬∙𝑺 = 𝒌𝑷𝑬 𝑷𝑷𝑬𝑪𝑽 −𝑪𝑷𝑬∙𝑺𝑲𝑷𝑬

𝒓𝑺 = 𝒌𝑺𝑪𝑷𝑬∙𝑺𝑪𝑯∙𝑺𝟐

𝒓𝑫𝑷A = 𝒌𝑷𝑨𝑪𝑷𝑨∙𝑺 − 𝒌−𝑷𝑨𝑷𝑷𝑨𝑪𝑽 = 𝒌𝑷𝑨 𝑪𝑷𝑨∙𝑺 −𝑷𝑷𝑨𝑪𝑽𝑲𝑫

Adsorption: 𝒓𝑨𝑯𝟐= 𝒌𝑯𝟐

𝑷𝑯𝟐𝑪𝑽𝟐 − 𝒌−𝑯𝟐

𝑪𝑯∙𝑺𝟐 = 𝒌𝑯𝟐

𝑷𝑯𝟐𝑪𝑽𝟐 −

𝑪𝑯∙𝑺𝟐

𝑲𝑯𝟐

𝐾𝑃𝐴 =1

𝐾𝐷

So the rate of desorption is 𝒓𝑫𝑷A = 𝒌𝑷𝑨 𝑪𝑷𝑨∙𝑺 −𝑲𝑷𝑨𝑷𝑷𝑨𝑪𝑽


42

Since the surface reaction is rate- determining

𝒓𝑨𝑷𝑬 = 𝒌𝑷𝑬 𝑷𝑷𝑬𝑪𝑽 −𝑪𝑷𝑬∙𝑺𝑲𝑷𝑬

𝒓𝑨𝑯𝟐= 𝒌𝑯𝟐

𝑷𝑯𝟐𝑪𝑽𝟐 −

𝑪𝑯∙𝑺𝟐

𝑲𝑯𝟐

𝒓𝑫𝑷𝑨 = 𝒌𝑷𝑨 𝑪𝑷𝑨∙𝑺 −𝑲𝑷𝑨𝑷𝑷𝑨𝑪𝑽

𝒓𝑨𝑷𝑬𝒌𝑷𝑬

≈ 𝟎 𝑪𝑷𝑬∙𝑺 = 𝑲𝑷𝑬𝑷𝑷𝑬𝑪𝑽

𝒓𝑫𝑷𝑨𝒌𝑷𝑨

≈ 𝟎 𝑪𝑷𝑨∙𝑺 = 𝑲𝑷𝑨𝑷𝑷𝑨𝑪𝑽

𝒓𝑨𝑯𝟐

𝒌𝑯𝟐

≈ 𝟎 𝑪𝑯∙𝑺 = 𝑲𝑯𝟐𝑷𝑯𝟐

𝑪𝑽

Substituting for 𝑪𝑷𝑬∙𝑺 and 𝑪𝑯∙𝑺 in the rate equation

𝒓𝑺 = 𝒌𝑺𝑪𝑷𝑬∙𝑺𝑪𝑯∙𝑺𝟐 = 𝒌𝑺𝑲𝑷𝑬𝑲𝑯𝟐

𝑷𝑷𝑬𝑷𝑯𝟐𝑪𝑽𝟑


43

𝐶𝑡 = 𝐶𝑉 + 𝐶𝑃𝐸∙𝑆 + 𝐶𝐻∙𝑆 + 𝐶𝑃𝐴∙𝑆

Substituting for 𝑪𝑷𝑬∙𝑺, 𝑪𝑯∙𝑺 and 𝑪𝑷𝑨∙𝑺 , factoring out 𝑪𝑽 :


𝐶𝑉 =𝐶𝑡

1 + 𝐾𝑃𝐸𝑃𝑃𝐸 + 𝐾𝐻2𝑃𝐻2 + 𝐾𝑃𝐴𝑃𝑃𝐴

So, the rate equation

−𝒓𝑨=𝒌𝑺𝑲𝑷𝑬𝑲𝑯𝟐

𝑷𝑷𝑬𝑷𝑯𝟐𝑪𝒕𝟑

𝟏 + 𝑲𝑷𝑬𝑷𝑷𝑬 + 𝑲𝑯𝟐𝑷𝑯𝟐 +𝑲𝑷𝑨𝑷𝑷𝑨𝟑

=𝒌𝑷𝑷𝑬𝑷𝑯𝟐

𝟏 + 𝑲𝑷𝑬𝑷𝑷𝑬 + 𝑲𝑯𝟐𝑷𝑯𝟐 +𝑲𝑷𝑨𝑷𝑷𝑨𝟑

Where: 𝑘 = 𝐶𝑡3𝑘𝑆𝐾𝑃𝐸𝐾𝐻2


44

−𝑟𝐴=𝑘𝑃𝑃𝐸𝑃𝐻2

1 + 𝐾𝐻2𝑃𝐻2 + 𝐾𝑃𝐸𝑃𝑃𝐸 + 𝐾𝑃𝐴𝑃𝑃𝐴3


1 + 𝐾𝑃𝐸𝑃𝑃𝐸 + 𝐾𝑃𝐴𝑃𝑃𝐴


1 + 𝐾𝐻2𝑃𝐻2 + 𝐾𝑃𝐸𝑃𝑃𝐸 + 𝐾𝑃𝐴𝑃𝑃𝐴2


2. Eley-Rideal (hydrogen reacts directly from gas-phase)

3. Langmuir-Hinshelwood, hydrogen adsorbs molecularly

Reaction rate laws (surface reaction RDS):

Homework: Derive rest

of rate laws!


45

From site balance: which

components occupy sites

From site balance: how

many sites are occupied

in RDS step

From RDS: which components

react in RDS

Square root refers to adsorption

that occupies two sites

−𝑟𝐴=𝑘′𝑃𝐴𝑃𝐵

1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐶𝑃𝐶2

It is known that the rate law for irreversible gas-phase reaction A + B2

→ C is

What is the mechanism behind this rate law?

From rate law to mechanism

46


−𝑟𝐴=𝑘′𝑃𝐴𝑃𝐵

1 + 𝐾𝐵𝑃𝐵 + 𝐾𝐶𝑃𝐶2

It is known that rate law for irreversible gas-phase reaction A + B2 → C

is

B2 (g) + 2 S ↔ 2 B∙S

A (g) + 2 B∙S ↔ C∙S + S

C∙S ↔ C + S

Thus, mechanism is

Adsorption of B

occupying two sites (dissociative adsorption)

Surface reaction, RDS.

where A reacts from gas phase

Desorption of C

47

Irreversible Surface-Reaction-Limited Rate Laws


48

runReaction rate

(mol/dm3 s)

Cyclohexanol

PA (atm)

Cyclohexene PB

(atm)

Water

PW (atm)

1 3.30E-05 1 1 1

2 7.05E-05 5 1 1

3 7.65E-06 10 1 1

4 1.83E-05 2 1 5

5 1.49E-05 2 1 10

6 1.36E-05 3 5 0

7 1.08E-05 3 10 0

8 2.18E-05 2 1 1

9 1.67E-05 3 1 0

10 5.24E-05 3 3 3

11 5.60E-06 5 5 5

Deducing rate law from experimental data

Cyclohexanol (A) → Cyclohexene (B) + Water (W)

OH

+ H2O

49

runReaction rate

(mol/dm3 s)

Cyclohexanol

PA (atm)

Cyclohexene PB

(atm)

Water

PW (atm)

1 3.30E-05 1 1 1

2 7.05E-05 5 1 1

3 7.65E-06 10 1 1

4 1.83E-05 2 1 5

5 1.49E-05 2 1 10

6 1.36E-05 3 5 0

7 1.08E-05 3 10 0

8 2.18E-05 2 1 1

9 1.67E-05 3 1 0

10 5.24E-05 3 3 3

11 5.60E-06 5 5 5

Dependence of reaction rate on cyclohexanol (runs 1, 2 and 3)

→ −𝒓𝑨~𝑷𝑨

𝟏 + 𝑲𝑨𝑷𝑨 +⋯

Explanation: When PA is small, -rA increases linearly with increasing PA.

When PA is high, rate becomes constant


50

runReaction rate

(mol/dm3 s)

Cyclohexanol

PA (atm)

Cyclohexene PB

(atm)

Water

PW (atm)

1 3.30E-05 1 1 1

2 7.05E-05 5 1 1

3 7.65E-06 10 1 1

4 1.83E-05 2 1 5

5 1.49E-05 2 1 10

6 1.36E-05 3 5 0

7 1.08E-05 3 10 0

8 2.18E-05 2 1 1

9 1.67E-05 3 1 0

10 5.24E-05 3 3 3

11 5.60E-06 5 5 5

Dependence of reaction rate on cyclohexene (runs 6, 7 and 9)

−𝒓𝑨~𝑷𝑨

𝟏 + 𝑲𝑨𝑷𝑨 +𝑲𝑩𝑷𝑩 +⋯

Explanation: When PB increases, -rA decreases linearly.


51

runReaction rate

(mol/dm3 s)

Cyclohexanol

PA (atm)

Cyclohexene PB

(atm)

Water

PW (atm)

1 3.30E-05 1 1 1

2 7.05E-05 5 1 1

3 7.65E-06 10 1 1

4 1.83E-05 2 1 5

5 1.49E-05 2 1 10

6 1.36E-05 3 5 0

7 1.08E-05 3 10 0

8 2.18E-05 2 1 1

9 1.67E-05 3 1 0

10 5.24E-05 3 3 3

11 5.60E-06 5 5 5

Dependence of reaction rate on water (runs 4, 5 and 8)


𝟏 + 𝑲𝑨𝑷𝑨 +𝑲𝑩𝑷𝑩 +𝑲𝑾𝑷𝑾

Explanation: When PW increases, -rA decreases linearly.


52

runReaction rate

(mol/dm3 s)

Cyclohexanol

PA (atm)

Cyclohexene PB

(atm)

Water

PW (atm)

1 3.30E-05 1 1 1

2 7.05E-05 5 1 1

3 7.65E-06 10 1 1

4 1.83E-05 2 1 5

5 1.49E-05 2 1 10

6 1.36E-05 3 5 0

7 1.08E-05 3 10 0

8 2.18E-05 2 1 1

9 1.67E-05 3 1 0

10 5.24E-05 3 3 3

11 5.60E-06 5 5 5

Dependence of reaction rate on PA = PB = PW = P (runs 1, 10 and 11)


𝟏 + 𝑲𝑨𝑷𝑨 + 𝑲𝑩𝑷𝑩 +𝑲𝑾𝑷𝑾𝟐

Explanation: When P is small, -rA increases

linearly with increasing P. When P is high,

rate becomes constant


53

Based on given data, rate law for cyclohexanol dehydration is

−𝑟𝐴 =𝑘𝑃𝐴

1 + 𝐾𝐴𝑃𝐴 + 𝐾𝐵𝑃𝐵 + 𝐾𝑊𝑃𝑊2

Thus mechanism is

A + S ↔ A∙S

A∙S + S → B∙S + W∙S

B∙S ↔ B + S

W∙S ↔ W + S desorption of water

adsorption of cyclohexanol (A)

surface reaction on two sites, RDS

desorption of cyclohexene (B)


54

● PSSH = Pseudo-Steady-State Hypothesis

Applied in reactions which contain active intermediate species (e.g.

reactions involving free radicals)

● Net formation rate of active intermediate species is zero:

𝑟𝑖∙𝑆∗ = 0

(Compare to r/k≈0)

● PSSH useful when

Two or more steps are rate-limiting

Some steps are irreversible

● See example in the course book:

Chapter 10.3.6 (4th ed.)

Chapter 10.3.5 (3rd ed.)

Rate laws derived from PSSH

55


Example: the gas-phase decomposition of azomethane

CH3 2N2→C2H6 + N2

Mechanism:

Reaction 1: CH3 2N2 + CH3 2N2𝑘1AZO∗

CH3 2N2 + CH3 2N2∗

Reaction 2: CH3 2N2 + CH3 2N2∗𝑘2AZO∗

CH3 2N2 + CH3 2N2

Reaction 3: CH3 2N2∗𝑘3AZO∗

C2H6 + N2

56


Because each of the reaction steps is elementary, the corresponding rate

laws for the active intermediate AZO* in reactions 1,2,3 are

𝒓𝟏𝐀𝐙𝐎∗ = 𝒌𝟏𝑪𝐀𝐙𝐎𝟐

𝒓𝟐𝐀𝐙𝐎∗ = −𝒌𝟐𝑪𝐀𝐙𝐎𝑪𝐀𝐙𝐎∗

𝒓𝟑𝐀𝐙𝐎∗ = −𝒌𝟑𝑪𝐀𝐙𝐎∗

The concentration of the active intermediate AZO* is not readily

measurable, so we will use the Pseudo-Steady-State-Hypothesis (PSSH)

to obtain a rate law in terms of measurable concentrations.

( Let 𝑘1 = 𝑘1AZO∗, 𝑘2 = 𝑘2AZO∗, 𝑘3 = 𝑘3AZO∗)

57


The rate of formation of product:

𝒓𝐂𝟐𝐇𝟔 = 𝒌𝟑𝑪𝐀𝐙𝐎∗

To find the concentration of the active intermediate AZO*, we set the net

rate of formation of AZO* equal to zero, 𝑟AZO∗ ≡ 0

𝒓𝐀𝐙𝐎∗ = 𝒓𝟏𝐀𝐙𝐎∗ + 𝒓𝟐𝐀𝐙𝐎∗ + 𝒓𝟑𝐀𝐙𝐎∗ = 𝟎

= 𝒌𝟏𝑪𝐀𝐙𝐎𝟐 − 𝒌𝟐𝑪𝐀𝐙𝐎𝑪𝐀𝐙𝐎∗ − 𝒌𝟑𝑪𝐀𝐙𝐎∗ = 𝟎

Solving for 𝑪𝐀𝐙𝐎∗: 𝑪𝐀𝐙𝐎∗ =𝒌𝟏𝑪𝐀𝐙𝐎

𝟐

𝒌𝟐𝑪𝐀𝐙𝐎 + 𝒌𝟑

Substituting for 𝑪𝐀𝐙𝐎∗ in the rate Equation: 𝒓𝐂𝟐𝐇𝟔 =𝒌𝟏𝒌𝟑𝑪𝑨𝒁𝑶

𝟐

𝒌𝟐𝑪𝑨𝒁𝑶 + 𝒌𝟑

58

1. Select mechanism

2. Make rate law for each step (steps are elementary)

3. Assume rate-limiting step (RDS)

– For RDS, k is small

– For others -r/k ≈ 0

4. Solve surface concentrations

5. Write site balance and solve Cv

6. Substitute surface concentrations and Cv into rate law

7. Substitute Kp (if overall reaction is irreversible)

8. Compare with data

9. Extract parameters

Reaction mechanisms – Summary

59

1. Quasi-Equilibrium Hypothesis

1. Propose an active intermediate.

2. Propose a mechanism.

3. Write rate laws.

4. Write rate of formation of product.

5. Write net rate of formation of the active intermediate

and use the PSSH.

6. Eliminate the concentration of the active intermediate

species in the rate laws by solving the simultaneous

equations developed in Steps 4 and 5

Reaction mechanisms – Summary

60

2. Pseudo-Steady-State Hypothesis

Reaction Engineering in Microelectronic Fabrication

Chemical Vapor Deposition

The mechanism of CVD is very similar to a heterogeneous catalytic

reaction. The reactant(s) adsorbs on the surface site and then reacts on

the surface to form a new surface site. This process may be followed by a

desorption step, depending on the particular reaction.

The growth of germanium films can be accomplished by CVD.

A proposed mechanism is:

For example:

61


GeCl4 (g) GeCl2 (g) + Cl2 (g)

GeCl2 (g) + S GeCl2 ∙ S𝑘𝐴

H2 (g) + 2S 2H ∙ S𝑘𝐻

Gas-phase dissociation:

Adsorption:

Adsorption:

Surface reaction: ?

S

GeCl2 ∙ S + 2H ∙ S Ge (S) + 2HCl (g) + 2S𝑘𝑆

62


The surface reaction is RDS

𝒓𝐃𝐞𝐩′′ = 𝒌𝑺𝒇𝐆𝐞𝐂𝐥𝟐𝒇𝑯

𝟐

Where

𝑟𝐷𝑒𝑝′′ = deposition rate per unit surface area, nm/s

𝑘𝑆 = surface specific reaction rate, nm/s

𝑓𝐺𝑒𝐶𝑙2 = fraction of the surface occupied by germanium dichloride

𝑓𝐻 = fraction of the surface covered by hydrogen atom

The difference between developing CVD rate laws and rate laws for

catalysis is that the site concentration is replaced by the fractional

surface area coverage.

Fractional area balance: 𝒇𝑽 + 𝒇𝑮𝒆𝑪𝒍𝟐 + 𝒇𝑯 = 𝟏

63

The net rate of GeCl2 adsorption is

𝒓𝑨𝐆𝐞𝐂𝐥𝟐 = 𝒌𝑨 𝑷𝐆𝐞𝐂𝐥𝟐𝒇𝑽 −𝒇𝐆𝐞𝐂𝐥𝟐𝑲𝑨

Since the surface reaction is rate-limiting, in a manner analogous to

catalysis reactions, we have for the adsorption of GeCl2 :

𝒓𝑨𝐆𝐞𝐂𝐥𝟐𝒌𝑨

≈ 𝟎

𝒇𝐆𝐞𝐂𝐥𝟐 = 𝑲𝑨𝑷𝐆𝐞𝐂𝐥𝟐𝒇𝑽

The fractional surface coverage of GeCl2 :


64

The dissociative adsorption of hydrogen


𝒓𝑨𝑯𝟐= 𝒌𝑯 𝑷𝑯𝟐

𝒇𝑽𝟐 −

𝒇𝑯𝟐

𝑲𝑯

𝒓𝑨𝑯𝟐

𝒌𝑯≈ 𝟎

Since the surface reaction is rate-limiting

Then 𝒇𝑯 = 𝒇𝑽 𝑲𝑯𝑷𝑯𝟐

65

𝒓𝐃𝐞𝐩′′ = 𝒌𝑺𝒇𝐆𝐞𝐂𝐥𝟐𝒇𝑯

𝟐 = 𝒇𝑽𝟑𝒌𝑺𝑲𝑨𝑷𝐆𝐞𝐂𝐥𝟐𝑲𝑯𝑷𝑯𝟐

Substituting for𝒇𝐆𝐞𝐂𝐥𝟐and𝒇𝑯 in the rate equation, we obtain:

We solve for 𝒇𝑽 in an identical manner to that for 𝑪𝑽 in heterogeneous

catalysis.

𝒇𝑽 + 𝒇𝑽 𝑲𝑯𝑷𝑯𝟐+ 𝒇𝑽𝑲𝑨𝑷𝐆𝐞𝐂𝐥𝟐 = 𝟏

Rearranging yields

𝒇𝑽 =𝟏

𝟏 + 𝑲𝑨𝑷𝐆𝐞𝐂𝐥𝟐 + 𝑲𝑯𝑷𝑯𝟐

66


Finally, substituting for 𝒇𝑽 in the rate equation

𝒓𝐃𝐞𝐩′′ =

𝒌𝑺𝑲𝑯𝑲𝑨𝑷𝐆𝐞𝐂𝐥𝟐𝑷𝑯𝟐


𝟑

and lumping𝑲𝑯, 𝑲𝑨, and 𝒌𝑺 into a specific reaction rate 𝒌′yields


𝒌′𝑷𝐆𝐞𝐂𝐥𝟐𝑷𝑯𝟐


𝟑

67

We need to relate the partial pressure of GeCl2 to the partial pressure of

GeCl4 in order to calculate the conversion of GeCl4. If we assume that

the gas-phase reaction

GeCl4 (g) GeCl2 (g) + Cl2 (g)

is in equilibrium, we have

𝐾𝑃 =𝑃GeCl2𝑃Cl2𝑃GeCl4

𝑃GeCl2 =𝑃GeCl4𝑃Cl2

∙ 𝐾𝑃

and if hydrogen is weakly adsorbed 𝑲𝑯𝑷𝑯𝟐< 𝟏 , we obtain the rate of

deposition as



𝒌𝑷𝐆𝐞𝐂𝐥𝟒𝑷𝑯𝟐𝑷𝐂𝐥𝟐𝟐

𝑷𝐂𝐥𝟐 +𝑲𝑷𝑷𝐆𝐞𝐂𝐥𝟒𝟑

68

69

Extended reading:

Surface chemkin: A general formalism and software

for analyzing heterogeneous chemical kinetics at a

gas‐surface interface, International Journal of

Chemical Kinetics, DOI:

https://doi.org/10.1002/kin.550231205

CatalyticReaction Engineering

Yongdan Li

Nov-Dec, 2018

Professor of Industrial ChemistryDepartment of Chemical and Metallurgical EngineeringSchool of Chemical TechnologyAalto UniversityEmail: [email protected] 1, E404

Date post:	06-Oct-2020
Category:	Documents
Upload:	others
View:	6 times
Download:	0 times

Catalytic Reaction Engineering - Aalto · Derive reaction rate law when propene (PE) adsorbs...

Documents