SOLID STATE AND SURFACE CHEMISTRY LECTURE 11 OBJECTIVES

SOLID STATE AND SURFACE CHEMISTRY (LECTURE 11)

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

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 Ka A(g) + S(surface) ⇄ A—S(surface) Kd Kc = Ka/Kd = [A—S]/[A][S]

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]

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.

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/k. BT If we define b = Kc/ KB T, Equation (6 ) becomes: ( kb is Boltzmann constant) 1/ θ = 1+ 1/b. PA 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.

LANGMUIR ADSORPTION ISOTHERM 1/ θ = 1+ 1/b. PA θ = b. PA/1+ b. PA 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 b. PA chemisorption and one-layer physisorption the Langmuir

EXAMPLE (1) The data given below are for the adsorption of CO on charcoal at 273 k. 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 (101. 325 k. Pa) P/k. Pa V/cm 3 13. 3 10. 2 26. 7 18. 6 40 25. 5 53. 3 31. 5 66. 7 36. 9 80 41. 6 93. 3 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 1/V 0. 0752 0. 0374 0. 025 0. 0191 0. 098039 0. 053763 0. 039216 0. 031746 0. 0149 0. 0125 0. 01072 0. 0271 0. 024038 0. 021692

A plot of 1/v versus 1/p will have a slope of 1. 128 and an intercept of 0. 009. and an intercept of 1/vm= 0. 009 → 0. 12 R 2 = 0. 9999 0. 1 vm = 111 cm 3 slope (1/bvm )= 1. 182 b = 1/1. 182× 111 1/V 0. 08 0. 06 0. 04 0. 02 0 0 b = 7. 6 × 10 -3 k. Pa-1 0. 02 0. 03 0. 04 1/P 0. 05 0. 06 0. 07 0. 08

EXAMPLE (2) Langmuir studied the adsorption of N 2 (g) onto a mica surface at 273. 15 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 m 3 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 10 -5 and an intercept of 0. 252 → vm = 3. 96 x 10 -8 m 3 b = 1/1. 18 × 10 -5 × 3. 96 x 10 -8 = 2. 14 × 10 12 Torr -1 At STP → 1 mole gas 22. 4 l = 0. 0224 m 3 ? mole gas = vm = 3. 96 x 10 -8 m 3 >>>>>> 1. 77 x 10 -6 mol № of molecules = 1. 77 x 10 -6 mol x 6. 022 x 10 23= 1. 06 x 10 18 σ◦ = 1. 06 x 10 18 / (0. 01 m)2 = 1. 06 x 10 22 m -2

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

θ 2 = kc (1 - θ )2 [A 2 ] θ = kc ½ [A 2 ] ½ - kc ½ [A 2 ] ½ θ θ + kc ½ [A 2 ] ½ θ = kc ½ [A 2 ] ½ θ(1 + kc ½ [A 2 ] ½ ) = kc ½ [A 2 ] ½ θ= kc ½ [A 2 ] ½ /(1 + kc ½ [A 2 ] ½ ) 1/θ= 1 + 1/kc ½ [A 2 ] 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= 1 + 1/b A 2 ½ PA 2 ½