Derivation of Bet equation and different isotherms

This file includes BET adsorption isotherm and five types of isotherms, derivation of BET adsorption isotherms.

1. BET Equation
By
Hari Bhakta Oli
Lecture Note 2019

2. BET theory aims to explain the physical adsorption of gas
molecules on a solid surface and serves as the basis for an
important analysis technique for the measurement of the
specific surface area of materials. In 1938, Stephen
Brunauer, Haul Hugh Emmett, and Edward Teller
published an article about the BET theory in a journal for
the first time.
The concept of the theory is an extension of the Langmuir theory, which is
a theory for monolayer molecular adsorption, to multilayer adsorption with
the following hypothesis

a) the gas molecules physically adsorb on a solid in layers
infinitely
b) Langmuir theory can be applied to each layer

3. ??????
??????
??????(??????
0−??????)
=
1
??????
????????????
+
??????−1
??????
????????????
.
??????
??????
0


Where, V
m and C are constants, a plot between
??????
??????
??????(??????
0−??????)
vs
??????
??????
0
gives straight line with solpe
??????−1
??????
????????????
and
intercept equal to
1
??????
????????????
.
If the value of C is: C>>1, then equation takes the
form
??????
??????
??????
??????
=
1
(1−
??????
??????
0
)

where, C = K
1/K
L = exp
(∆desH
1-
∆vapH
L
)

4. When, C>>1 and P/P
0 <<1 ; adsorption is monolayer
∆desH
1>>
∆vapH
L
Example: monolayer formation of N
2 at -195℃ and O
2
at -183℃ on charcoal surface
P 
x/m


Adsorbent surface
Monolayer adsorption
Type I Isotherm

5. When, C>1, formation of multilayer takes only
after compete formation of monolayer
∆desH
1>
∆vapH
L
Example: adsorption of N
2 at-195℃ on Fe catalyst and on
silica gel
P 
x/m


Adsorbent surface
Monolayer
Ad layer
Type II Isotherm

6. When, C<1,
∆desH
1<
∆vapH
L , formation of
multilayer takes place from very beginning
Example: Adsorption of Br
2 and I
2 on
silica gel at 79℃
P 
x/m


Adsorbent surface
Multilayer formation
Type III Isotherm

7. Formation of multilayer takes place along with the
possibility of filling the capillary pores as a result of
condensation of the adsorbate at pressure
appreciably below the saturation vapour pressure.
Adsorption of C
6H
6 in Fe
2O
3 at 50℃
P 
x/m


Pores and capillaries
Type IV Isotherm

8. Multilayer formation from very beginning. Filling
on capillary and pores.
Adsorption of H
2O in charcoal at 100℃
P 
x/m


Multilayer formation in Pores
and capillaries
Type V Isotherm

9. The formation of multilayer may be represented by the
following equation
??????+??????
&#3627408498;
&#3627409359;
??????−??????
??????−??????+??????
&#3627408498;
&#3627409360;
??????
&#3627409360;−??????
??????
&#3627409360;−??????+??????
&#3627408498;
&#3627409361;
??????
&#3627409361;−??????
??????
&#3627408527;−&#3627409359;−??????+??????
&#3627408498;
&#3627408527;
??????
&#3627408527;−??????
G = un adsorbed gases molecule
S = vacant site of adsorbent
GS = surface single molecule is adsorbed
G
2S = double molecule adsorbed per vacant site
&#3627408498;
&#3627409359;=
[????????????]
??????[??????]
&#3627408498;
&#3627409360;=
[??????
&#3627409360;??????]
??????−??????[??????]
&#3627408498;
&#3627409361;=
[??????
&#3627409361;??????]
??????
&#3627409360;−??????[??????]

Derivation

10. We know,
[G]α[P] (pressure of gas)
[S]α[θ
0] (the fraction of free surface)
[GS] α[θ
1] (fraction of surface covered with single
molecule adsorption)
[G
2S] α[θ
1] (fraction of surface covered with two
molecule adsorption)
Therefore,
&#3627408498;
&#3627409359;=
??????&#3627409359;
????????????&#3627409358;
,&#3627408498;
&#3627409360;=
??????&#3627409360;
????????????&#3627409359;
,&#3627408498;
&#3627409361;=
??????&#3627409361;
????????????&#3627409360;
and so o….(1)
The value of constant K
1 is usually very large as
compare to rest of equilibrium constants. It is
because the interaction between the adsorbate and
the adsorbent decreases very rapidly as the distance
from the surface is increased.

11. The remaining constants K
2, K
3,… etc though will not have
the same values but the differences between any two
constants is generally much smaller than that between K
1
and K
2. so it is assumed that,
&#3627408498;
&#3627409360;≅&#3627408498;
&#3627409361;≅&#3627408498;
&#3627409362;≅⋯………≅&#3627408498;
&#3627408499;………..(&#3627409360;)
where, K
L is the equilibrium constant corresponding to the
saturated vapour liquid equilibrium and is given by
&#3627408498;
&#3627408499;=
&#3627409359;
??????
&#3627409358;
…………….(3)
now, form equation (1), (2) and (3)
??????
&#3627409359;=&#3627408498;
&#3627409359;????????????
&#3627409358;
??????
&#3627409360;=&#3627408498;
&#3627409360;????????????
&#3627409359;=
&#3627409359;
??????
&#3627409358;
??????&#3627408498;
&#3627409359;????????????
&#3627409358;=&#3627408498;
&#3627409359;????????????
&#3627409358;
??????
??????
&#3627409358;

??????
&#3627409361;=&#3627408498;
&#3627409361;????????????
&#3627409360;= &#3627408498;
&#3627409359;????????????
&#3627409358;
??????
??????
&#3627409358;
&#3627409360;
and so on

12. If we assume that the entire adsorbent surface is covered
then total coverage of the first layer will be given by
θ
total= θ
0+ θ
1+ θ
2 + θ
3+…….. = 1
θ
0 +&#3627408498;
&#3627409359;????????????
&#3627409358;+&#3627408498;
&#3627409359;????????????
&#3627409358;
??????
??????
&#3627409358;
+ &#3627408498;
&#3627409359;????????????
&#3627409358;
??????
??????
&#3627409358;
&#3627409360;
+ …… =1
θ
0[1+&#3627408498;
&#3627409359;??????(1+
??????
??????
&#3627409358;
+
??????
??????
&#3627409358;
&#3627409360;
+….)]=1
The solution for small value of P/P
0 is
1+
??????
??????
&#3627409358;
+
??????
??????
&#3627409358;
&#3627409360;
+…=[1-
??????
??????
&#3627409358;
]
-1
= 1/(1-
??????
??????
&#3627409358;
)
Therefore,
θ
0[1+
&#3627408498;
&#3627409359;??????
(1−
??????
??????
&#3627409358;
]=1
or, ??????
&#3627409358;=
&#3627409359;−
??????
??????
&#3627409358;
&#3627409359;+&#3627408498;
&#3627409359;??????−
??????
??????
&#3627409358;
……………..(4)

13. Total volume of adsorbed gas is given by
??????
??????&#3627408528;????????????&#3627408525;=??????
&#3627408526;&#3627408528;&#3627408527;&#3627408528;??????
&#3627409359;+&#3627409360;??????
&#3627409360;+&#3627409361;??????
&#3627409361;+ ….
or, ??????
??????&#3627408528;????????????&#3627408525;=??????
&#3627408526;&#3627408528;&#3627408527;&#3627408528;. [&#3627408498;
&#3627409359;????????????
&#3627409358;(1+2
??????
??????
&#3627409358;
+ &#3627409361;
??????
??????
&#3627409358;
&#3627409360;
+ … ]
since, (1+2
??????
??????
&#3627409358;
+ &#3627409361;
??????
??????
&#3627409358;
&#3627409360;
+ …)=
&#3627409359;
&#3627409359;−
??????
??????
&#3627409358;
&#3627409360;
Substituting this in above equation we get,
??????
??????=??????
&#3627408526;
&#3627408498;
&#3627409359;????????????
&#3627409358;
&#3627409359;−
??????
??????
&#3627409358;
&#3627409360; (from equation 4)
??????
??????=??????
&#3627408526;
&#3627408498;
&#3627409359;??????
&#3627409359;−
??????
??????
&#3627409358;
&#3627409360;
&#3627409359;−
??????
??????
&#3627409358;
&#3627409359;+&#3627408498;
&#3627409359;??????−
??????
??????
&#3627409358;

??????
??????=??????
&#3627408526;
&#3627408498;
&#3627409359;??????
&#3627409359;−
??????
??????
&#3627409358;
(&#3627409359;+&#3627408498;
&#3627409359;??????−
??????
??????
&#3627409358;
)
…….(5)

14. Substituting the value of P as
??????= ??????
0
??????
??????
0
=
1
??????
??????
??????
??????
0

??????
??????=??????
&#3627408526;
&#3627408498;
&#3627409359;
??????
??????
??????
??????
0

&#3627409359;−
??????
??????
&#3627409358;
(&#3627409359;+
&#3627408498;
&#3627409359;
??????
??????
??????
??????
0
−
??????
??????
&#3627409358;
)

??????
??????=??????
&#3627408526;
??????
&#3627408526;??????.??????.
??????
&#3627409358;−??????[&#3627409359;+(C−1)
??????
??????
&#3627409358;
]

15. ??????
??????
??????(??????
&#3627409358;−??????)
=
&#3627409359;
??????
????????????
+
??????−&#3627409359;
??????
????????????
.
??????
??????
&#3627409358;

Where, V
m and C are constants, a plot between
??????
????????????(??????0−??????)
vs
??????
??????0
gives straight
line with solpe
??????−1
??????
????????????
and intercept equal to
1
??????
????????????
.

16. Thank You