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Derivation of Bet equation and different 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. 𝑃 𝑉𝑡(𝑃0 − 𝑃) = 1 𝑉𝑚 𝐶 + 𝐶 − 1 𝑉𝑚 𝐶 . 𝑃 𝑃0 Where, Vm and C are constants, a plot between 𝑃 𝑉𝑡(𝑃0−𝑃) vs 𝑃 𝑃0 gives straight line with solpe 𝐶−1 𝑉 𝑚 𝐶 and intercept equal to 1 𝑉 𝑚 𝐶 . If the value of C is: C>>1, then equation takes the form 𝑉𝑡 𝑉𝑚 = 1 (1 − 𝑃 𝑃0 ) where, C = K1/KL = exp(∆desH1-∆vapHL)
- 4. When, C>>1 and P/P0 <<1 ; adsorption is monolayer ∆desH1>>∆vapHL Example: monolayer formation of N2 at -195℃ and O2 at -183℃ on charcoal surface P x/m Adsorbent surface Monolayeradsorption Type I Isotherm
- 5. When, C>1, formation of multilayer takes only after compete formation of monolayer ∆desH1>∆vapHL Example: adsorption of N2 at-195℃ on Fe catalyst and on silica gel P x/m Adsorbent surface Monolayer Ad layer Type II Isotherm
- 6. When, C<1, ∆desH1<∆vapHL , formation of multilayer takes place from very beginning Example: Adsorption of Br2 and I2 on silica gel at 79℃ P x/m Adsorbent surface Multilayerformation 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 C6H6 in Fe2O3 at 50℃ P x/m Pores and capillaries Type IV Isotherm
- 8. Multilayer formationfrom very beginning. Filling on capillary and pores. Adsorption of H2O in charcoal at 100℃ P x/m Multilayerformation in Pores and capillaries Type V Isotherm
- 9. The formation of multilayer may be represented by the following equation 𝐆 + 𝐒 𝑲 𝟏 𝑮 − 𝑺 𝐆 − 𝐒 + 𝐒 𝑲 𝟐 𝑮 𝟐 − 𝑺 𝑮 𝟐 − 𝑺 + 𝐒 𝑲 𝟑 𝑮 𝟑 − 𝑺 𝑮 𝒏−𝟏 − 𝑺 + 𝐒 𝑲 𝒏 𝑮 𝒏 − 𝑺 G = un adsorbed gases molecule S = vacant site of adsorbent GS = surface single molecule is adsorbed G2S = double molecule adsorbed per vacant site 𝑲 𝟏 = [𝑮𝑺] 𝑮 [𝑺] 𝑲 𝟐 = [𝑮 𝟐 𝑺] 𝑮 − 𝑺 [𝑺] 𝑲 𝟑 = [𝑮 𝟑 𝑺] 𝑮 𝟐 − 𝑺 [𝑺] Derivation
- 10. We know, [G]α[P] (pressure of gas) [S]α[θ0] (the fraction of free surface) [GS] α[θ1] (fraction of surface covered with single molecule adsorption) [G2S] α[θ1] (fraction of surface covered with two molecule adsorption) Therefore, 𝑲 𝟏 = 𝜽 𝟏 𝑷𝜽 𝟎 , 𝑲 𝟐 = 𝜽 𝟐 𝑷𝜽 𝟏 , 𝑲 𝟑 = 𝜽 𝟑 𝑷𝜽 𝟐 and so o….(1) The value of constant K1 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 K2, K3,… etc though will not have the same values but the differences between any two constants is generally much smaller than that between K1 and K2. so it is assumed that, 𝑲 𝟐 ≅ 𝑲 𝟑 ≅ 𝑲 𝟒 ≅ ⋯ … … … ≅ 𝑲 𝑳 … … … . . (𝟐) where, KL is the equilibrium constant corresponding to the saturated vapour liquid equilibrium and is given by 𝑲 𝑳 = 𝟏 𝑷 𝟎 …………….(3) now, form equation (1), (2) and (3) 𝜽 𝟏 = 𝑲 𝟏 𝑷𝜽 𝟎 𝜽 𝟐 = 𝑲 𝟐 𝑷𝜽 𝟏 = 𝟏 𝑷 𝟎 𝑷 𝑲 𝟏 𝑷𝜽 𝟎 = 𝑲 𝟏 𝑷𝜽 𝟎 𝑷 𝑷 𝟎 𝜽 𝟑 = 𝑲 𝟑 𝑷𝜽 𝟐 = 𝑲 𝟏 𝑷𝜽 𝟎 𝑷 𝑷 𝟎 𝟐 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 θtotal= θ0+ θ1+ θ2 + θ3+…….. = 1 θ0 +𝑲 𝟏 𝑷𝜽 𝟎+𝑲 𝟏 𝑷𝜽 𝟎 𝑷 𝑷 𝟎 + 𝑲 𝟏 𝑷𝜽 𝟎 𝑷 𝑷 𝟎 𝟐 + …… =1 θ0[1+𝑲 𝟏 𝑷(1+ 𝑷 𝑷 𝟎 + 𝑷 𝑷 𝟎 𝟐 +….)]=1 The solution for small value of P/P0 is 1+ 𝑷 𝑷 𝟎 + 𝑷 𝑷 𝟎 𝟐 +…=[1- 𝑷 𝑷 𝟎 ]-1= 1/(1- 𝑷 𝑷 𝟎 ) Therefore, θ0[1+ 𝑲 𝟏 𝑷 (1− 𝑷 𝑷 𝟎 ]=1 or, 𝜽 𝟎 = 𝟏− 𝑷 𝑷 𝟎 𝟏+𝑲 𝟏 𝑷− 𝑷 𝑷 𝟎 ……………..(4)
- 13. Total volume of adsorbed gas is given by 𝑽 𝒕𝒐𝒕𝒂𝒍 = 𝑽 𝒎𝒐𝒏𝒐 𝜽 𝟏 + 𝟐𝜽 𝟐 + 𝟑𝜽 𝟑 + … . or, 𝑽 𝒕𝒐𝒕𝒂𝒍 = 𝑽 𝒎𝒐𝒏𝒐. [𝑲 𝟏 𝑷𝜽 𝟎(1+2 𝑷 𝑷 𝟎 + 𝟑 𝑷 𝑷 𝟎 𝟐 + … ] since, (1+2 𝑷 𝑷 𝟎 + 𝟑 𝑷 𝑷 𝟎 𝟐 + … ) = 𝟏 𝟏− 𝑷 𝑷 𝟎 𝟐 Substituting this in above equation we get, 𝑽 𝒕 = 𝑽 𝒎 𝑲 𝟏 𝑷𝜽 𝟎 𝟏− 𝑷 𝑷 𝟎 𝟐 (from equation 4) 𝑽 𝒕 = 𝑽 𝒎 𝑲 𝟏 𝑷 𝟏− 𝑷 𝑷 𝟎 𝟐 𝟏− 𝑷 𝑷 𝟎 𝟏+𝑲 𝟏 𝑷− 𝑷 𝑷 𝟎 𝑽 𝒕 = 𝑽 𝒎 𝑲 𝟏 𝑷 𝟏− 𝑷 𝑷 𝟎 (𝟏+𝑲 𝟏 𝑷− 𝑷 𝑷 𝟎 ) …….(5)
- 14. Substituting the value of P as 𝑃 = 𝑃0 𝑃 𝑃0 = 1 𝐾𝐿 𝑃 𝑃0 𝑽 𝒕 = 𝑽 𝒎 𝑲 𝟏 𝐾𝐿 𝑃 𝑃0 𝟏 − 𝑷 𝑷 𝟎 (𝟏 + 𝑲 𝟏 𝐾𝐿 𝑃 𝑃0 − 𝑷 𝑷 𝟎 ) 𝑽 𝒕 = 𝑽 𝒎 𝑽 𝒎 𝑪. 𝑷. 𝑷 𝟎 − 𝑷 [𝟏 + (C−1) 𝑷 𝑷 𝟎 ]
- 15. 𝐏 𝐕𝐭(𝐏𝟎 − 𝐏) = 𝟏 𝐕 𝐦 𝐂 + 𝐂 − 𝟏 𝐕 𝐦 𝐂 . 𝐏 𝐏𝟎 Where, Vm and C are constants, a plot between 𝑃 𝑉𝑡(𝑃0−𝑃) vs 𝑃 𝑃0 gives straight line with solpe 𝐶−1 𝑉 𝑚 𝐶 and intercept equal to 1 𝑉 𝑚 𝐶 .
- 16. Thank You