1. Solid State and Surface Chemistry (lecture 11)

2. Objectives By the end of this section you should:
Be able to drive Langmuir isotherm
Know the different form of Langmuir equation
Know the physical meaning of Langmuir isotherm

3. أنواع أيزوثيرمات الإمتزازعند درجة حرارة ثابتةI II
III IV V
relative pres. P/P0
1.0
amount adsorbed
تم العثور على خمسة أنواع من أيزوثرم الإمتزاز الفيزيائي physisorption على جميع المواد الصلبة: :Type Iو يوجد في المواد المسامية التي لها مسامات صغيرة مثل الفحم. و من الواضح أن هذا النوع يتبع أيزوثيرم لانجمير أحادي قديتبع الادمصاص الكيميائي))الطبقة. Type II: للمواد غير المسامية non-porous materials :Type III للمواد المسامية التي يسهل اختراقها مع قوة تماسك بين جزيئات المواد الممتزة أكبر من قوة التلاصق بين جزيئات المادة الممتزة و سطح المادة المازة. Type IV: امتزاز متعدد المراحل حيث تتكون طبقة إمتزاز أحادية الطبقة و من ثم يتم بناء طبقات إضافية. Type V: للمواد المسامية التي يسهل اختراقها مع قوة تلاصق بين جزيئات المواد الممتزة و سطح المادة المازة أكبر من قوة التماسك بين جزيئات المادة الممتزة.

4. The dynamic equilibrium is Kc = Ka/Kd = [A—S]/[A][S]
Langmuir isotherm
This simplest expression for an adsorption isotherm was first derived by Irving Langmuir in 1918.
Its based on 3 assumptions:
monolayer adsorption (so no multilayer adsorption)
no interaction between adsorbed molecules
homogeneous surface(all adsorption sites energetically identical)
The dynamic equilibrium is
A(g) + S(surface) ⇄ A—S(surface)
Kc = Ka/Kd = [A—S]/[A][S] Ka Kd

5. Langmuir adsorption isotherm
Let σо be the conc. of surface sites in units of m-2.
θ the fraction of surface sites occupied by an adsorbate.
σ the adsorbate conc. on the surface = θ σо
The conc. of empty surface = (1- θ ) σо
[A] is conc. of A(g) , PA is pressure of A(g),
kb Boltzman constant
Rate of desorption = υd = kd θ σо
Rate of adsorption = υa = ka (1- θ ) σо [A]
At equilibrium
υd = υa
kd θ σо = ka (1- θ ) σо [A]

6. At very high concentrations 1-θ = 1 / Kc [A]
θ σо = ka/ kd (1- θ ) σо [A] Kc= ka/kd θ = Kc [A] - θ Kc [A] θ + θ Kc [A] = Kc [A] θ (1+ Kc [A]) = Kc [A] θ = Kc [A]/ (1+ Kc [A]) 1/θ = 1/ Kc [A]/ + Kc [A]/ Kc [A] 1/θ = 1+ 1/ Kc [A]
At very high concentrations
1-θ = 1 / Kc [A]
i.e. The surface becomes saturated with molecules at high pressures.

7. ( kb is Boltzmann constant)
For the adsorption of gases on solids the concentration of the gas [A] can be replaced by the partial pressure of the gas P.
So, the ideal gas law can be used, then [A] = PA/kBT
If we define b = Kc/ KB T, Equation (6 ) becomes:
( kb is Boltzmann constant)
1/ θ = 1+ 1/bPA
Let vm is the amount that can be adsorbed at monolayer coverage and v the amount adsorbed at any pressure (for a unit mass of adsorbent) then:
θ = v / vm
we can rewrite the last two equations as:
1/v = 1/ Pbvm +1/vm
From the equation, we see that a plot of 1/v versus 1/p will have a slope of 1/bvm and an intercept of 1/vm.

8. Langmuir adsorption isotherm
1/ θ = 1+ 1/bPA
θ = bPA/1+ bPA
PA = 1/b (θ /1-θ)
θ approaches unity due to the adsorption of a monolayer on the surface as the P become large θ
θ and b can be calculated from Langmuir adsorption isotherm
bPA
chemisorption and one-layer physisorption the Langmuir

9. EXAMPLE (1)
The data given below are for the adsorption of CO on charcoal at 273k. Confirm that they fit the langmuir isotherm and fit the langmuir isotherm, and find the constant k and the volume corresponding to complete coverage. In each case V has been corrected to a atm ( kPa)
P/kPa
13.3
26.7 40
53.3
66.7 80
93.3
V/cm3
10.2
18.6
25.5
31.5
36.9
41.6
46.1
1/v = 1/ Pbvm +1/vm
A plot of 1/v versus 1/p will have a slope of 1/bvm and an intercept of 1/vm.
1/P
0.0752
0.0374
0.025
0.0191
0.0149
0.0125
1/V
0.0271

10. A plot of 1/v versus 1/p will have a slope of 1
A plot of 1/v versus 1/p will have a slope of and an intercept of
and an intercept of 1/vm= →
vm = 111 cm3
slope (1/bvm )= 1.182
b = 1/1.182×111
b = 7.6 × kPa-1

11. EXAMPLE (2)
Langmuir studied the adsorption of N2 (g) onto a mica surface at K. From the data presented below, determine the values of b and vm , the volume of gas that corresponds to a monolayer coverage. Use this value of vm to determine the total number of surface sites.
P/10-12 Torr
2.55
1.79
1.30
0.98
0.71
0.46
0.30
0.21
V/ 10-8 m3
3.39
3.17
2.89
2.62
2.45
1.95
1.55
1.23
A plot of 1/v versus 1/p will have a slope of 1.18 x and an intercept of
intercept of → vm = 3.96 x m3
b = 1/1.18 ×10-5 × 3.96 x = 2.14 × Torr -1
At STP → 1 mole gas 22.4 l = m3
?mole gas = vm = 3.96 x m3 >>>>>> 1.77 x mol
№ of molecules = 1.77 x mol x x 10 23= 1.06 x 10 18
σ◦ = 1.06 x / (0.01m)2 = 1.06 x m -2

12. EXAMPLE (2)
Derive the Langmuir adsorption isotherm for the case in which a diatomic molecule dissociates upon adsorption to the surface.
A2(g) + S(surface) ⇄ A—S(surface)
Kc = Ka/Kd = [A—S]2/[A2][S]2
Rate of desorption = υd = kd θ2 σо2
Rate of adsorption = υa = ka (1- θ )2 σо 2[A2]
At equilibrium
υd = υa
kd θ σо 2 = ka (1- θ )2 σо 2[A2]
θ 2 = ka (1- θ )2 [A2]/ kd Ka Kd

13. θ= kc ½ [A2 ] ½ /(1 + kc ½ [A2 ] ½ )
For the adsorption of gases on solids the concentration of the gas [A] can be replaced by the partial pressure of the gas P.
1/θ= /b A2 ½ PA2 ½