13.1 Catalysis
567
This theory was devised by Irving
Langmuir, 1881–1957, an American
industrial chemist who won the 1932
Nobel Prize in chemistry for his work on
surface chemistry.
Let θ represent the fraction of the surface sites that are occupied by adsorbed
A molecules. The adsorption is assumed to be an elementary process so that the rate of
adsorption is proportional to the concentration of A in the fluid phase and to 1 − θ, the
fraction of surface sites available for adsorption:
rate of adsorption k 1 [A](1 − θ)
(13.1-5)
where [A] is the concentration of A in the gas or liquid phase and where k 1 is a rate
constant that can depend on temperature but not on [A] or θ. The desorption is also
assumed to be an elementary process so that
rate of desorption k
1 θ
(13.1-6)
where k
1 is another rate constant. At equilibrium, the rate of desorption equals the rate
of adsorption:
k
1 θ k 1 [A](1 − θ)
(13.1-7)
This equation can be solved for θ to give the Langmuir isotherm:
θ
k 1 [A]
k
1 + k 1 [A]
K[A]
1 + K[A]
(Langmuir isotherm)
(13.1-8)
where K is an equilibrium constant given by
K
k 1
k
1
(13.1-9)
The name “isotherm” is used because the formula corresponds to a fixed temperature. The value of K can be determined from a graph of the Langmuir isotherm since
1/K is equal to the value of [A] that corresponds to θ 1/2. Figure 13.2 schematically
depicts the Langmuir isotherm for a hypothetical system.
Exercise 13.2
Show that 1/K equals the value of [A] that corresponds to θ 1/2.
1
1
2
0 [A] = 1/K
[A]
Figure 13.2 The Langmuir Isotherm.
567
This theory was devised by Irving
Langmuir, 1881–1957, an American
industrial chemist who won the 1932
Nobel Prize in chemistry for his work on
surface chemistry.
Let θ represent the fraction of the surface sites that are occupied by adsorbed
A molecules. The adsorption is assumed to be an elementary process so that the rate of
adsorption is proportional to the concentration of A in the fluid phase and to 1 − θ, the
fraction of surface sites available for adsorption:
rate of adsorption k 1 [A](1 − θ)
(13.1-5)
where [A] is the concentration of A in the gas or liquid phase and where k 1 is a rate
constant that can depend on temperature but not on [A] or θ. The desorption is also
assumed to be an elementary process so that
rate of desorption k
1 θ
(13.1-6)
where k
1 is another rate constant. At equilibrium, the rate of desorption equals the rate
of adsorption:
k
1 θ k 1 [A](1 − θ)
(13.1-7)
This equation can be solved for θ to give the Langmuir isotherm:
θ
k 1 [A]
k
1 + k 1 [A]
K[A]
1 + K[A]
(Langmuir isotherm)
(13.1-8)
where K is an equilibrium constant given by
K
k 1
k
1
(13.1-9)
The name “isotherm” is used because the formula corresponds to a fixed temperature. The value of K can be determined from a graph of the Langmuir isotherm since
1/K is equal to the value of [A] that corresponds to θ 1/2. Figure 13.2 schematically
depicts the Langmuir isotherm for a hypothetical system.
Exercise 13.2
Show that 1/K equals the value of [A] that corresponds to θ 1/2.
1
1
2
0 [A] = 1/K
[A]
Figure 13.2 The Langmuir Isotherm.