Examples: Adsorption
Example 1: Using the BET Isotherm (adapted from Atkins, Physical Chemistry)
In an experiment to estimate the surface area of nutrile (TiO 2 ) by following the
adsorption of nitrogen gas, the following data were obtained at 75K. Confirm that the
data fit the BET isotherm in the range of pressures reported and hence determine the
surface area of 5g of substrate.
p/Torr
V/cm3
1.20
601
14.0
720
45.8
822
87.5
935
127.7
1046
164.4
1146
204.7
1254
Data:
At 75K, P o = 570 Torr
The radius of a molecule can be taken as 112.8 µm
All volumes have been corrected to 1 atm and 273K and refer to 1g of substrate.
Solution
The BET Isotherm
(C − 1)P
P
1
=
+
V (Po − P ) V m C V m CPo
Plot
P
P
vs
Po
V ( Po − P )
⇒ slope =
( C − 1)
VmC
and y-intercept =
1
VmC
Hence generate the following table
p/Torr
1.2
14.0
45.8
87.5
127.7
164.4
204.7
V/cm3
601
720
822
935
1046
1146
1254
0.002
0.025
0.080
0.154
0.224
0.288
0.359
3.51E-06
3.50E-05
1.06E-04
1.94E-04
2.76E-04
3.54E-04
4.47E-04
p/p o
p/[V(p o -p)]
Plot of P/[V(Po-P)] vs P/Po
5.00E-04
4.50E-04
4.00E-04
P/[V(Po-P)]
3.50E-04
y = 1.23E-03x + 4.06E-06
3.00E-04
2.50E-04
2.00E-04
1.50E-04
1.00E-04
5.00E-05
0.00E+00
0.00
0.05
0.10
0.15
0.25
0.20
0.30
P/Po
From the graph
•
Y-intercept:
•
Slope:
1
= 4.06 x 10−6 cm −3
VmC
( C − 1) = 1.23 x 10−3 cm −3
VmC
Solving the equations gives:
1.23 x 10−3
C −1 =
= 303
4.06 x 10−3
Vm =
⇒C =
304
1
= 813 cm3
−3
304 x 4.06 x 10
PV
At 1 atm and 273 K:=
n =
RT
(1.01 x 10 ) 813 x 10=
5
8.314 x 273
−6
0.036 mol
But there are 6.022 x 1023 molecules/mole.
Hence, we have 0.036 x 6.022 x 1023 = 2.18 x 1022 molecules
The surface area of each molecule = 4πr2 = 0.16 x 10-18 m2
Surface area/g = 3488 m2
For 5 g = 17440 m2
0.35
0.40
Example 2: Using the Langmuir Isotherm (from Atkins, Physical Chemistry)
The data below are for the adsorption of CO on charcoal at 273K. 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 1atm.
p/Torr
V/cm3
100
10.2
200
18.6
300
25.5
400
31.5
500
36.9
Solution strategy
The Langmuir isotherm is
θ=
where θ =
Kp
1 + Kp
(1)
V
→ V m = volume corresponding to complete coverage
Vm
Rearrange (1): ⇒ Kpθ + θ =
Kp
⇒
Hence plot
⇒ Kp
V V
+
=
Kp
Vm Vm
p P
1
=
+
V Vm KVm
p
vs p
V
⇒ slope =
1
1
and y-intercept =
Vm
KVm
600
41.6
700
46.1
Example 3: Determination of the enthalpy of adsorption (Atkins, Physical Chemistry)
The data below show the pressures of CO needed for the volume of adsorption (corrected
to 1 atm and 273 K) to be 10.0 cm3 using the same sample as in Example 2 above.
Calculate the adsorption enthalpy at this surface coverage.
T/K
P/Torr
200
30.0
210
37.1
220
45.2
230
54.0
240
63.5
Solution Strategy
The Langmuir isotherm is θ =
⇒ ln K + ln p= ln
Kp
1 + Kp
θ
⇒ Kp =
1−θ
θ
= const
1−θ
o
∆H ads
 ∂ ln K 
=


RT 2
 ∂T θ
o
∆H ads
 ∂ ln P 
 ∂ ln K 
⇒
=
−
=
−



RT 2
 ∂T θ
 ∂T θ
1
o d
∆
H
∂
ln
P


ads
T
⇒
 =
R dT
 ∂T θ


o
 ∂ ln P 
∆H ads
⇒
=
1 
R
 ∂

 T θ
o
∆H ads
Plot ln P vs 1/T and slope =
R
1
1
⇐ T =
− 2
dT
T
d
250
73.9