13.1 Catalysis
569
B 0.203 g −1
K
0.203 g −1
1.55 × 10 −4 mol L −1 g −1
1310 L mol −1
The asymptotic value of m is equal to 1/B, or 4.93 g.
b. The effective area of the sample is obtained from the assumption that the asymptotic
amount adsorbed corresponds to a full monolayer. Since the molar mass of chloroethane
is equal to 64.515 g mol −1 , the asymptotic amount adsorbed is equal to 0.0764 mol. The
effective area A is
A
0.260 nm 2 molecule −1
6.022 × 10 23 molecule mol −1
(0.0764 mol)
1.20 × 10 22 nm 2 1.20 × 10 4 m 2
Although this area corresponds to the area of a macroscopic square 110 m on a side,
charcoal can be so finely divided that this sample of charcoal might have a mass of only
a few grams.
There are two classes of adsorption processes: physical adsorption and chemical
adsorption (chemisorption). In physical adsorption, the binding forces are London
dispersion forces, dipole–dipole attractions, and so on. In chemical adsorption covalent
chemical bonds are formed between the atoms or molecules of the surface and the atoms
or molecules of the adsorbed substance. The Langmuir isotherm applies to both classes
if only a monolayer of atoms or molecules can be adsorbed on the surface and if the
adsorbed molecules do not dissociate. There are other isotherms that apply to the case
of multiple layers. 3
When hydrogen molecules are adsorbed on platinum they dissociate into hydrogen
atoms that are covalently bonded (chemically adsorbed) to platinum atoms on the
surface. In this case a different isotherm from that of Eq. (13.1-8) applies. Assume that
a substance A 2 dissociates to form two adsorbed A atoms that occupy two sites on the
surface:
A 2 (g) + 2 surface sites 2A(adsorbed)
(13.1-12)
If the adsorption is an elementary process,
rate of adsorption k 1 [A 2 ](1 − θ)
2
(13.1-13)
If desorption is an elementary process,
rate of desorption k
1 θ
2
(13.1-14)
The rates of adsorption and desorption are equated and the resulting equation is solved
for θ to obtain the equilibrium isotherm:
θ
K 1/2 [A 2 ] 1/2
1 + K 1/2 [A 2 ] 1/2 (adsorption with dissociation)
(13.1-15)
3 K. J. Laidler, op. cit., p. 234 (note 1).
569
B 0.203 g −1
K
0.203 g −1
1.55 × 10 −4 mol L −1 g −1
1310 L mol −1
The asymptotic value of m is equal to 1/B, or 4.93 g.
b. The effective area of the sample is obtained from the assumption that the asymptotic
amount adsorbed corresponds to a full monolayer. Since the molar mass of chloroethane
is equal to 64.515 g mol −1 , the asymptotic amount adsorbed is equal to 0.0764 mol. The
effective area A is
A
0.260 nm 2 molecule −1
6.022 × 10 23 molecule mol −1
(0.0764 mol)
1.20 × 10 22 nm 2 1.20 × 10 4 m 2
Although this area corresponds to the area of a macroscopic square 110 m on a side,
charcoal can be so finely divided that this sample of charcoal might have a mass of only
a few grams.
There are two classes of adsorption processes: physical adsorption and chemical
adsorption (chemisorption). In physical adsorption, the binding forces are London
dispersion forces, dipole–dipole attractions, and so on. In chemical adsorption covalent
chemical bonds are formed between the atoms or molecules of the surface and the atoms
or molecules of the adsorbed substance. The Langmuir isotherm applies to both classes
if only a monolayer of atoms or molecules can be adsorbed on the surface and if the
adsorbed molecules do not dissociate. There are other isotherms that apply to the case
of multiple layers. 3
When hydrogen molecules are adsorbed on platinum they dissociate into hydrogen
atoms that are covalently bonded (chemically adsorbed) to platinum atoms on the
surface. In this case a different isotherm from that of Eq. (13.1-8) applies. Assume that
a substance A 2 dissociates to form two adsorbed A atoms that occupy two sites on the
surface:
A 2 (g) + 2 surface sites 2A(adsorbed)
(13.1-12)
If the adsorption is an elementary process,
rate of adsorption k 1 [A 2 ](1 − θ)
2
(13.1-13)
If desorption is an elementary process,
rate of desorption k
1 θ
2
(13.1-14)
The rates of adsorption and desorption are equated and the resulting equation is solved
for θ to obtain the equilibrium isotherm:
θ
K 1/2 [A 2 ] 1/2
1 + K 1/2 [A 2 ] 1/2 (adsorption with dissociation)
(13.1-15)
3 K. J. Laidler, op. cit., p. 234 (note 1).