Adsorption - Isotherm Theory(Adsorption Consider activated...

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Adsorption 1 Isotherm Theory (Adsorption) Consider activated carbon (95% pass 200 mesh) one cubic foot of this material in water can present a surface area for adsorption of about 3,000,000 square feet. Consider a particle: Consider ion or molecules or something in solution “ “ to be adsorbed: Concentration concentration of ’s to be adsorbed “‘s are free to adsorb and desorb depending on concentration gradients: Rate at which particles are being adsorbed is proportional to the concentration in solution and the number of unoccupied sites, where θ is the fraction unoccupied. R adsorb C * β (1 − θ) (1) R adsorb = k 1 β C (1 - θ) (2) We assume that the rate at which particles are being desorbed is proportional to the number of occupied sites. R desorb βθ (3) o r Rk desorb = 2 βθ (4) When equilibrium is attained, there will be no net change, dR dt = 0 , and the rate of adsorption = rate of desorption ( ) = kC k e 12 1 β θ βθ (5) β ’s cancel & kC k C k e 11 2 = θ θ (6) Solving for θ , θ= + kk C e e 1 21 (7) but the fraction of sites occupied at equilibrium per unit is proportional to the weight of material adsorbed per unit weight of adsorbent Adsorption sites available (may be surface area)
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Adsorption 2 o r θ∝ X M (8) where X = weight of substance adsorbed M = w e i g h t o f a d s o r b e n t ( C A R B O N ) ∴= = + θ k X M kC kk C e e 3 1 21 (9) rearranging, k X M k k C k k C e e 3 1 2 1 2 1 = + let k b 3 1 = and k k a 1 2 = then: Langmuir X M abC aC e e = + 1 (10) straight line form M Xb a b C e =+ ⋅ 111 text claims that b = amount adsorbed to form a monolayer Example: Take 4 - 500 mL of water, add 1, 3, 15 and 30 mg of activated carbon to the beakers. Pipet 1 mL containing the equivalent of 50 µ g phenol into each beaker and stir 90 minutes. Test for phenol in solution at equilibrium. Results: #1 40 µ g/L, #2 16 µ g/L, #3 10 µ g/L and #4 4 µ g/L of phenol. Does it fit the Langmuir isotherm and what are the constants? m 1 = 1 mg X 1 50 40 2 30 =− = M X = 1 0 033 . 1 0 025 C e  = . m 2 = 3 mg X 2 50 16 2 42 =−= M X = 2 0 071 . 1 0 0625 C e  = . m 3 = 15 mg X 3 50 10 2 45 M X = 3 0 333 . 1 0100 C e  = . m 4 = 30 mg X 4 50 4 2 48 = M X = 4 0 625 . 1 025 C e  = .
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Adsorption 3 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 0.05 0.15 0.25 (1/Ce) 1/b = -0.05 b = -20 slope = 0.3/0.1 = 3 = 1/ab a = 3/-20 = -0.15 Freundlich Isotherm X M kC e n = 1 same units straight line form: Log X M Logk n LogC e =+ 1 1.0 10 100 1.0 10 100 1/n C e X/M Log k
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Adsorption 4 Adsorption a) usually limited to activated carbon because of economics b) Removes dissolved solids by mostly physical attraction (could be chemisorption sometimes) at a interface - here a solid/liquid interface but could be L/L, G/L, G/S, or as is in this case, S/L.
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This note was uploaded on 01/14/2012 for the course ENV 6015 taught by Professor Taylor during the Fall '11 term at University of Central Florida.

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Adsorption - Isotherm Theory(Adsorption Consider activated...

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