it is very important

1. Today: The Kinetics of Heterogeneous Catalysis, cont’d
Adsorption Isotherms
Rate Equations
“What adsorption properties must be balanced for an effective catalyst?”
ChemE 2200 – Chemical Kinetics Lecture
16
Recap: The Basic Mechanism of Heterogeneous Catalysis
The Langmuir-Hinshelwood Mechanism
of Elementary Reactions:
A  P
M
adsorption
A
M
M
A
A
ads,
A
des,
-
k
k



reaction
surface
P
M
A
M
surface
-
-
k

desorption
M
P
P
M
P
des,
P
ads,



k
k
-
usually the RLS
pre-equilibrium
post-equilibrium

2. The Thermodynamics of
Adsorption
Often in equilibrium.
will ignore
in ChemE 2200
A
M
M
A
A
ads,
A
des,
-
k
k



depends on adsorbate, surface, and coverage.
ads
H

Two Categories of Adsorption:
Physical Adsorption (aka physisorption)
Weak M-A bond: ads
H

 is 0 to 25 kJ/mol.
Examples: saturated hydrocarbons: CH4, CH3CH3
Chemical Adsorption (aka chemisorption)
Strong M-A bond: ads
H

 is 40 kJ/mol and higher
Examples: dissociative adsorption: H2 + 2M 2M-H


polar compounds: CO, NO, H2O
Almost always exothermic.

3. The Thermodynamics of Adsorption,
cont’d
good catalyst: comparable adsorption
poor catalysts:
adsorbed H2
excludes
adsorbed N2
poor catalysts:
adsorbed O2 excludes
adsorbed CO
good catalyst:
comparable adsorption
Comparable adsorption is necessary, but not sufficient.
For the catalyzed reaction A + B
M
 P the surface reaction is
M-A + M-B  M + M-P
Both A and B must be adsorbed on M.

4. The Thermodynamics of Adsorption,
cont’d
Adsorption of both reactants is essential
to the Langmuir-Hinshelwood Mechanism.
An alternative reaction,
M-A + B(gas)  M-P
is unlikely and rare.
Ballistic impact from the gas phase
is the Eley-Rideal Mechanism.
Like a termolecular elementary reaction,
A(gas) + B(gas) + C(gas)  P
reaction by ballistic impact is possible
but improbable
relative to the reaction of two adsorbed reactants.

5. The Thermodynamics of Adsorption - Adsorption
Isotherms constant T
A
PA
need A = f(PA)
M M M M M M M
M M M M M M M
M M M M M M M
A
A A
A
A A A A
A
A
need [M]  …, [M-A]  …
excellent model for
PA < vapor pressure of A.
sites
adsorption
total
A
by
occupied
sites
adsorption
A
of
Coverage
Fractional A 


Langmuir Isotherm.
Assumes ads
H
 is independent of surface coverage.
Assumes only a single layer of adsorbed A.
A
M
M
A
A
ads,
A
des,
-
k
k



At equilibrium: ]
A
M
[
]
M
][
A
[ A
des,
A
ads, -
k
k 

6. KA[A](1A)  A
0
A
A
des,
0
A
A
ads, ]
M
[
]
M
)[
1
](
A
[ 


 k
k
The Thermodynamics of Adsorption - Adsorption
Isotherms
from the definition of A:
substitute
[A] is
awkward
for
reactor
design.
A
M
M
A
A
ads,
A
des,
-
k
k



need [M]  …, [M-A]  …
At equilibrium: ]
A
M
[
]
M
][
A
[ A
des,
A
ads, -
k
k 
total concentration of adsorption sites  [M]0  [M] + [M-A]
0
A
]
M
[
]
A
M
[ -

  [M-A]  A[M]0
0
0
A
]
M
[
]
M
[
]
M
[ 

  [M]  (1A)[M]0
A
des,
A
ads,
A
Define
k
k
K 
]
A
[
1
A]
[
A
A
A
K
K



Convert [A] to PA:
RT
P
V
n A
A
A]
[ 

A
A
A
A
A
1 P
RT
K
P
RT
K




7. The Thermodynamics of Adsorption - Adsorption
Isotherms
Using the Langmuir Isotherm in Rate Equations.
Need expressions for [M-A] and [M-P] to derive the rate equation.
Assume Langmuir Isotherms for the adsorption/desorption of A and P.
This assumption allows one to write 
kads, A[M][A]  kdes, A[M-A] and kads, P[M][P]  kdes, P[M-P]
A Langmuir-Hinshelwood
Mechanism of
Elementary Reactions:
This assumption yields two useful expressions:
[M-A]  Kads, A[M][A] and [M-P]  Kads, P[M][P]
Overall Reaction: (gas)
M
(gas) P
A 
A
M
M
A
A
ads,
A
des,
-
k
k


 adsorption
surface reaction
P
M
P
M
P
des,
P
,
ads



k
k
- desorption
P
M
A
M
surface
-
-
k


8. Rate Equations for Heterogeneously-Catalyzed
Reactions
Recall our goal for Chemical Kinetics:
Given the Overall Reaction and Rate Equation,
devise a Mechanism of Elementary Reactions.
Recall our Strategy:
Gain experience by starting with
a Mechanism of Elementary Reactions
and deriving the Rate Equation.

9. Heterogeneous Catalysis: Case Study
1 metal catalyst
RLS
pre-equilibrium
post-equilibrium
Overall Reaction: (gas)
M
(gas) P
A 
A Langmuir-Hinshelwood
Mechanism of
Elementary Reactions:
A
M
M
A
A
ads,
A
des,
-
k
k


 adsorption
surface reaction
P
M
P
M
P
des,
P
,
ads



k
k
- desorption
P
M
A
M
surface
-
-
k


10. 0
P
ads,
A
ads,
A
ads,
]
M
[
]
P
[
]
A
[
1
]
A
[
]
A
M
[
K
K
K
-



0
P
ads,
A
ads,
A
ads,
]
M
[
]
M
][
P
[
]
M
][
A
[
]
M
[
]
M
][
A
[
K
K
K



Heterogeneous Catalysis: Case Study 1 - Example
1
Because the surface reaction is the RLS:
instead, jump to this
unoccupied M sites M-A sites M-P sites
Overall Reaction: (gas)
M
(gas) P
A 
RLS
pre-equilibrium
post-equilibrium
]
A
M
[
]
P
[
s -
k
dt
d

0
A ]
M
[
θ
]
A
M
[ 
- 0
]
M
[
]
P
M
[
]
A
M
[
]
M
[
]
A
M
[
-
-
-



Assume Langmuir Isotherms for A and P.
]
A
M
[
]
P
[
s -
k
dt
d
 0
P
ads,
A
ads,
A
ads,
s
]
M
[
]
P
[
]
A
[
1
]
A
[
K
K
K
k




11. Heterogeneous Catalysis: Case Study 1 - Example
2
RLS
pre-equilibrium
post-equilibrium
0
ads 
G
okay?
0
and
because
0
have
may
c),
(exothermi
0
although
ads
ads
ads
ads
ads
ads











S
S
T
H
G
G
H
A and P are
weakly adsorbed.
surface sites are
mostly unoccupied.
RLS 0 0
clue in rate equation:
denominator  1
[M] » [M-A], [M-P]  1 » Kads,A, Kads, P
]
A
M
[
]
P
[
s -
k
dt
d
 0
P
ads,
A
ads,
A
ads,
s
]
M
[
]
P
[
]
A
[
1
]
A
[
K
K
K
k


 0
A
ads,
s ]
M
][
A
[
K
k

assume
Langmuir isotherms

12. 0
P
ads,
A
ads,
A
ads,
s
]
M
[
]
P
[
]
A
[
1
]
A
[
K
K
K
k



Heterogeneous Catalysis: Case Study 1 - Example
3
RLS
pre-equilibrium post-equilibrium
A is strongly adsorbed.
P is weakly adsorbed.
surface sites are
mostly occupied by A.
0 0
Rate is independent of [A]! How can this be?
The RLS is proportional to [M-A]. Increasing
[A] does not increase [M-A] because the
surface is saturated with adsorbed A.
Clues in rate equation:
1. denominator = 1
2. No [A] term.
[M-A] » [M], [M-P]  Kads,A » 1, Kads, P
assume
Langmuir isotherms
RLS
]
A
M
[
]
P
[
s -
k
dt
d
 0
s ]
M
[
k


13. ]
A
M
[
]
P
[
s -
k
dt
d
 0
P
ads,
A
ads,
A
ads,
s
]
M
[
]
P
[
]
A
[
1
]
A
[
K
K
K
k



Heterogeneous Catalysis: Case Study 1 - Example
4
RLS
pre-equilibrium
post-equilibrium
A is weakly adsorbed.
P is strongly adsorbed.
surface sites are
mostly occupied by P.
0 0
The product P in an irreversible reaction
hinders the reaction! How can this be?
P monopolizes the surface sites,
which reduces [M-A].
clue in rate equation:
denominator contains
only a [P] term.
[M-P] » [M], [M-A]  Kads,P » 1, Kads, A
assume
Langmuir isotherms
RLS
0
P
ads,
A
ads,
s
]
M
[
]
P
[
]
A
[
K
K
k


14. 0
empty ]
M
[
θ
]
M
[ 
0
P
ads,
P
ads,
s
s
]
M
[
]
M
][
P
[
]
M
][
P
[
]
M
[
]
M
[
]
M
[
K
K
k
k- 


]
M
][
P
[
]
P
M
[ P
ads,
K
- 
0
]
M
[
]
P
M
[
]
A
M
[
]
M
[
]
M
[
-
- 


Heterogeneous Catalysis: Case Study 1 - Example
5
RLS
post-equilibrium post-equilibrium
RLS:
by definition
RLS
clues in rate equation: denominator contains ‘unoccupied sites” term and [P] term.
]
M
][
A
[
]
P
[
A
ads,
k
dt
d

Assume Langmuir Isotherm for P.
Assume post-equilibrium for surface reaction: ]
P
M
[
]
A
M
[ s
s -
k
-
k -
 ]
M
][
P
[
P
ads,
sK
k-

0
s
s
P
ads,
]
M
[
]
P
[
1
1
1











k
k
K -
]
M
][
A
[
]
P
[
A
ads,
k
dt
d
 0
s
s
P
ads,
A
ads,
]
M
[
]
P
[
1
1
]
A
[











k
k
K
k
-