Chap. 22. Processes on Solid Surfaces

Chap. 22. Processes on Solid Surfaces

Catalyst

accelerates a reaction

undergoes no net chemical changes

Homogeneous Catalyst: A catalyst in the same phase as the reaction mixture, e.g., the decomposition of

hydrogen peroxide in aqueous solution catalyzed by iodide ion.

Heterogeneous Catalyst: A catalyst in a different phase from the reaction mixture, e.g., hydrogenation of

ethene (C₂H₄) to ethane (C₂H₆), a gas phase

Thin-Film Deposition

- - - -

Figure 22C.1

A catalyst provides a different path with a lower activation

enthalpy. The result is an increase in the rate of formation of products.

G

Surface Characteristics of Silicon (001)

~Fig. 22A.1

~Fig. 22A.2 Terrace

~Fig. 22A.4 STM

~Fig. 22A.5 STM

~Fig. 22A.6 AFM

Seoul National University ⁴ Joonhyeon Kang et al.

Strongly Adsorbed Weakly Adsorbed

Catalyst: Altered Activation Barrier

U. Stimming, J. Electrochem. Soc. (2005).

T = 300 K

Pt

Au

Hydrogen chemisorption energy

Free energy for hydrogen adsorption

http://bp.snu.ac.kr

~Fig. 22C.2

Log (Exchange

Current)

Log (Exchange

Current)

Volcano Curve

22A.2. Physisorption and Chemisorption

Physisorption:

- the enthalpy of physisorption

Chemisorption:

- the enthalpy of chemisorption

- usually,

∆H <0, ∆S<0

20 kJ mol

≈ −

200 kJ mol

≈ −

Assumptions

1. Adsorption cannot proceed beyond monolayer coverage.

______

_______

- - - - (skip)

Types of Adsorption Modes

Physical Adsorption or

Physisorption

Chemical Adsorption or

Chemisorption

Bonding between

molecules and surface is by

weak van der Waals forces.

Chemical bond is formed between molecules and surface.

Dr. K. L. Yeung (Hong Kong Univ.) http://teaching.ust.hk/~ceng511/notes/

6

Heterogeneous Catalysis

22B. Adsorption & Desorption

The fractional coverage θ : the extent of surface coverage

adsorption of

rate the

:

coverage monolayer

complete

to ing correspond adsorbate

of volume the

:

adsorbed adsorbate

of volume the

:

available sites

adsorption of

number

occupied sites

adsorption of

number

d V V

V V

θ θ θ

∞

∞

=

=

(22B.1a)

22B. The Dynamic Equilibrium

The rate of adsorption

The rate of desorption

At equilibrium, there is no net change.

Langmuir isotherm

) AM(surface

)

surface (

M )

(

A g + ←→^ka

kd

sites vacant

of number the

: ) - 1 (

sites adsorption

of number total

the :

A of pressure partial

the :

) 1

(

θ N

N p pN dt k

d

a

θ

θ

θ θ

N dt k

d

=

−

d a d

a

k K k

p K

p K

N k pN

k

+ =

=

∴

=

−

1

) 1

(

θ

θ θ

(22B.1a)

(22B.1b)

(22.B2)

______

_____

Θ : Fractional Coverage

(skip---)

Figure 22B.2

The Langmuir isotherm for

non-dissociative adsorption for different values of K.

(skip---)

Example 1

The data given below are the adsorption of CO on charcoal at 273 K. Confirm that they fit the Langmuir isotherm, and find the

constant K and the volume corresponding to complete coverage. In each case V has been corrected to 1.00 atm (101.325 kPa).

p/kPa 13.3 26.7 40.0 53.3 66.7 80.0 93.3 V/cm³ 10.2 18.6 25.5 31.5 36.9 41.6 46.1 Solution

with 1

Kp Kp V V

p p

V V KV

θ θ θ _∞

∞ ∞

+ = =

= +

Hence, a plot of p/V against p should give a straight line of slope and intercept

1 V_∞ 1 KV_∞.

p/kPa 13.3 26.7 40.0 53.3 66.7 80.0 93.3 (p/kPa)/(V/cm³) 1.30 1.44 1.57 1.69 1.81 1.92 2.02 The slope is 0.00900 and so The intercept at p=0 is 1.20, so

111 cm .3

V_∞ =

3 1

3 3

1 7.51 10 kPa

(111 cm ) (1.20 kPa cm )

K = ₋ = × ^{− −}

×

(skip---)

Figure 22B.1

The plot of the data in

Example. As illustrated here, the Langmuir isotherm

predicts that a straight line should be obtained when p/V is plotted against p.

(skip---)

The rate of adsorption for adsorption with dissociation is proportional to the pressure and to the probability that both atoms will find sites, which is proportional to the square of the number of vacant sites,

The rate of desorption

At equilibrium

Langmuir isotherm for

adsorption with dissociation

{

^θ

}

θ

N p dt k

d

a

( ) ^θ

θ

dt d

= d

−

{ } ( )

Kp Kp

N k N

p

k_{a d}

= +

∴

=

−

1

) 1

( ^{2 2}

θ

θ

θ

Figure 22B.3

The Langmuir isotherm for dissociative adsorption, for different values of K.

) surface (

2X )

(

X₂ g →

The isosteric standard enthalpy of adsorption

standard enthalpy of adsorption at a fixed surface coverage the van’t Hoff equation

Example 2:

The data below show the pressures needed for the volume of adsorption (corrected to 1.00 atm and 273 K) to be 10.0 cm³ using the same sample as in Example 1. Calculate the

adsorption enthalpy at this surface coverage.

T/K 200 210 220 230 240 250

p/kPa 4.00 4.95 6.03 7.20 8.47 9.85

2 0

ln ads

RT H T

K ∆

∂ =

∂

Solution:

The Langmuir isotherm can be rearranged to

Therefore, a plot of ln p against 1/T should be a straight line R

H T

p

T dT

T d

RT H T

K T

p p K

Kp

0 ads

2

2 0 ads

) / 1 ( ln

, / 1 /

) / 1 ( with

ln ln

constant ln

ln

constant is

when 1

= ∆



 





∂

∂

−

=

− ∆

 =



 





∂

− ∂

 =



 





∂

∴ ∂

= +

= −

θ

θ θ

θ θ

θ

The slope is −0.904, so

Figure 22B.4

The isoteric enthalpy of

adsorption can be obtained

from the slope of the plot of lnp against 1/T, where p is the

pressure needed to achieve the specified coverage.

( )

1 - 3 0

ads

mol kJ

52 . 7

10 904

. 0

−

=

×

×

−

=

∆H K R

Isotherm with multilayer adsorption

BET isothem:

{ }

( ^{H H} ) ^RT e

c c

V p

p z p

z c z

cz V

V

/ mon

mon

0 vap 0

constant :

coverage monolayer

to ing correspond volume

the :

liquid bulk

pure a

resembling

adsorbate, layer

a above pressure

vapor the

:

*

*

) 1

( 1 ) 1

(

∆

−

=

=

−

−

= −

Figure 22B.5

Plot of the BET isotherm for different values of c . The value of V /V_mon rises indefinitely

because the adsorbate may condense on the covered substrate surface.

Example 3:

The data below relate the adsorption of N₂ on rutile (TiO₂) at 75 K.

Confirm that they fit a BET isotherm in the range of pressures reported, and determine V_mon and c.

p/kPa 0.160 1.87 6.11 11.67 17.02 21.92 27.29

V/mm³ 601 720 822 935 1046 1146 1254

At 75 K, p*=76.0 kPa. The volumes have been corrected to 1.00 atm and 273 K and refer to 1.00 g of substrate.

Solution:

The BET equation can be reorganized into

(c−1)/cV can be obtained from the slope of a plot of the expression

mon mon

) 1 (

1 )

1

(

z c

cV V

z

z

−

+

− =

p/kPa 0.160 1.87 6.11 11.67 17.02 21.92 27.29

10³z 2.11 24.6 80.4 154 224 288 359

10⁴z/(1−z)(V/mm³) 0.035 0.350 1.06 1.95 2.76 3.53 4.47

When 10⁴z/(1−z) is plotted against 10³z, the intercept is 0.0398, so

The slope of the line is

3 - 6

mon

mm 10

98 .

1 3 ₋

× cV =

( )

3 mon

3 - 3

3 - 4

3 2

mon

mm 811

and 310

mm 10

23 . 1 mm

10 10

10 23

. 1 1

=

=

∴

×

=

×

×

×

− = _{− − −}

V c

cV c

so , 10 23 .

1 × ⁻²

Figure 22B.6

The BET isotherm can be tested, and the parameters determined, by plotting z/(1−z)V against z=p/p*.

Deviations from the Langmuir isotherm which assumes the independence and equivalence of the adsorption sites

The Temkin isotherm

1

( )

ln

, : constants c c p c c

θ

The Freundlich isotherm

2

1

1

c pc

θ

the BET isotherm

(skip)

z V

V c

= −

≈

1 1 ,

1

mon

22C. Homogeneous Catalysis

• Features of homogeneous catalysis

[ ]

+ 3 2

+ +

3 2 2 3 2 2 +

2 2 3

+ +

3 2 2 3 2

+

H O +H O H O +H O H O

H O H O

H O +I HOI+H O H O I

HOI+H O H O +O +I (fast)

a

K

v k

− −

−

 

 

=  

   

→ =    

→



- - - -

(skip---)

2000.

of factor a

by increases constant

rate The

ion iodide of

presence in the

mol kJ

57

catalyst of

absence in the

mol kJ

76

) ( O )

( O H 2 )

( O H 2

1 -

1 -

2 2

2 2

=

=

+

→

a a

E E

g l

aq

The second step is rate-determining step, therefore we can obtain the rate law of the overall reaction by setting the overall rate

equal to the rate of the second step:

[ ]

[

]

r a

O H O H O I

where .

d k

dt

k k K

   −

=    

=

In acid catalysis, the crucial step is the transfer of a proton to the substrate:

In base catalysis, a proton is transferred from the substrate to a base:

X + HA → HX⁺ + A⁻ HX⁺ → products

XH + →B X⁻ + BH⁺ X⁻ → products

(skip---)

(ES): active combination of the enzyme and substrate

Experiments show that the rate of formation of the product depends on the concentration of the enzyme, and so although the net reaction is simply S→P, this must reflect an underlying mechanism with

steps that involve the enzyme.

The Michaelis-Menten Mechanism

Enzymes

E P

S

E + → +

E P

ES)

( S

E +

+

k 'a →^kb

[ ] [ _{( )} ]

[ ( ) ^ES ] [ ][ ] ^{E S '} [ ( ) ^ES ] [ ( ) ^ES ]

P ES

b

k k

d k dt k d

−

−

=

=

Figure

The basis of the Michaelis-

Menten mechanism of enzyme action. Only a fragment of the large enzyme molecule E is shown.

S

E

(ES)

E P

The steady-state approximation

If [E]₀ is the total concentration of the enzyme, then [E]+[(ES)]=[E]₀. Since only a little enzyme is added,

[ ][ ] [ ( ) ] [ ( ) ] [ ( ) ] [ ][ ]

a b

a

b a

a

k k

k

k k

k

' S ES E

0 ES

ES '

S E

≈ +

∴

≈

−

−

[ ] [ ( ) ] [ ]

[ ( ) ] { [ ] [ ( ) ] } [ ]

[ ] [ ]

'

S ES

ES E

S ES

S

0

a b

a

k k

k

⇓

+

≈ −

∴

≈

+

The rate of enzymolysis depends linearly on the amount of enzyme added, and also on the amount of substrate present.

[ ] [ _{( )} ]

[ ( ) ] [ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

constant Michaelis

the ' is

where

S S E S

'

S E P

S '

S ES E

P ES

0 0

0

a a b

M

M a a

a b

b a

a a

b a b

k k k k

k k k

k k

k k dt

d

k k

k k dt k

d

= +

= + +

≈ +

∴

+

≈ +

=

[ ] [ ] [ ]

[ ] [ ] ^S

1 1

S S

, P E

0

b M b

M b

k k k

k

k k k

dt k d

+

=

= +

=

Figure

The variation of the effective rate constant k with substrate

concentration according to the Michaelis-Menten mechanism.

0 2 8

0.2 0.4 0.6 0.8 1.0

k/k b

4 6 10

0

[ ] [ ] [ ]

[ ] [ ]

[ ] [ ] [ ]

[ ] [ ] ^S

1 1

/ S 1

/ S

S S

S E S E

P

M

M M 1 0

0

b b

b M

b M b

k k k

k

k k k

k k

k k

k k k dt

d

+

=

= +

= +

+ ≅

=

Figure

A Lineweaver-Burk plot for the analysis of an enzymolysis that proceeds by a Michaelis-

Menten mechanism, and the significance of the intercepts and the slope.

[ ]

1 1

b b S k k = k + k

1/[S]

0

1/k_b

−1/K_M 1/k

Slope = K_M/k_b

(skip)

Problems from Chap. 22

D 22A.2 (AFM / SEM / STM / TEM)

D 22C.2 (Fig. 22B.8, Fig. 22C.2, and ppt 22-4)