Determine the isotherm that fits the data and give the constants of the equation using the given units.

Transcription

1 CHEE 3 Solved examples - Adsorption Problem : The data given below are for the orption of CO on charcoal at 273 K. Describ the orption using an appropriate orption isotherm and calculate the corresponding Gibbs energy of orption. p / kpa V / cm Problem 2: Batch tests were performed in the laboratory using solutions of phenol in water and particles of granular activated carbong. The quilibrium data at room temperature are shown in the table below as the dependence of phenol surface concentration on the equilibrium concentration in the solution. Determine the isotherm that fits the data. c / kg m Γ / kg kg Problem 3: Equilibrium isotherm data for orption of glucose from an aqueous solution to activated alumina are as follows: c / g cm Γ / g solute g alumina Determine the isotherm that fits the data and give the constants of the equation using the given units. Problem 4: Interaction of serum and plasma proteins with implanted biomaterials plays one of the major roles in the biocompatibility of the biomaterial. This is especially significant in the first few hours after the implantation, during which biomaterial-protein interactions occur. Adsorption of a protein layer onto medical implants in the body can lead to adverse effects such as inflammation, clot formation, and increase corrosion of metallic implants. Hence, it is of the major importance to characterize orption of major blood/serum proteins onto biomaterials surfaces. (a) Table below lists equilibrium orption data for the orption of serum protein fibrinogen onto a medical-grade stainless steel surface (36LVM) at 33 K. Your task is to use the Langmuir isotherm to describe this interaction, and subsequently to calculate the Gibbs energy of orption. Evaluate a suitability of the Langmuir isotherm by examining both the Γ f(c) and θ f(c) relationships. (b) You are also required to determine the orientation of the protein on the surface (side-on, end-on, or tilted and at what angle) at % coverage. The dimensions of the protein molecule are 6.5 nm 47.5 nm, while the molecular weight is 34 g/mol. (c) As mentioned above, the first few hours after the implantation are crucial due to the orption of serum proteins on the biomaterial s surface. An experiment was perfmored at 33 K with 36LVM coupons ( cm 2 ) in which the dependence of the surface amount (concentration) of fibrinogen was measured with time, and the corresponding data is presented in Table 2. The experiment was done using an advanced version of a FTIR technique, and the table shows the dependence of the integrated intensity of Amide I peak on time. Amide I peak is an IR fingerprint of proteins, and the listed integrated intensity values (A amide I / cm - ) are direclty proportional to the corresponding surface concentration (Γ / mg m -2 ) of the protein as:

2 A amid I.4379 Γ. Your task is to apply a first-order kinetic orption model to fit the data in Table 2, and to calculate the corresponding orption and desorption constants, the equilibrium constant, and the Gibss energy of orption. Compre the latter value to the Gibss energy of orption you calculated in (a), and comment the difference (if any). How long it would take to completely cover the implant s surface under the conditions applied? To solve the problem, you can use either a Solver function in Excel, Mathlab or any other similar software package. Figure : Fibrinogen molecule dimensions: 6.5 nm 47.5 nm Table. Dependence of the fibrinogen surface concentration on the corresponding equilibrium concentration in the bulk solution. c fib / g L Γ / mg m c fib / g L Γ / mg m Table 2. Dependence of the integrated intensity on time for the orption of fibrinogen on 36LVM from a solution containing. g/l of fibrinogen.. Time / min A amide I / cm

3 SOLUTIONS Problem : Using the Langmuir isotherm, p V BV mono + p V Plot p/v vs. p y.9x R p/v /kpacm p /kpa In order to determine ΔG, we first evaluate V mono from the slope of the curve: V mono. cm 3. Then from the intercept of the curve we calculate B.75 kpa -.76 atm -. Now ΔG is: B ΔG P exp o RT ΔG.76 exp Δ G 623. J/mol

4 Problem 2: log(c). Using the Langmuir isotherm, plot c/γ vs. c. Also using the Freundlich isotherm plot log(γ) vs c/γ /kg kg-.5 y 6.489x +. R c /kg m-3 log (Γ) y.284x R log (c) The Freundlich isotherm shows a better fit (R 2.994) than the Langmuir isotherm (R ).

5 Problem 3: log(c). Using the Langmuir isotherm, plot c/γ vs. c. Also using the Freundlich isotherm plot log(γ) vs..6.4 c/γ /g g y 7.83x +.75 R c /g cm-3 log (Γ) y.3898x R log (c) It is obvious from the plots that the data fits the Langmuir isotherm (R 2.999) better than the Freundlich isotherm (R 2.899). From the slope of the curve: Γ mono.4 g solute g - alumina. From the intercept B 6.5 cm 3 g - solute.

6 Problem 4: (a) Using the Langmuir isotherm, plot c/γ vs. c c/γ /m- 2 5 y x + 64 R c /g m-3 In order to determine ΔG, we first evaluate Γ mono from the slope of the curve: Γ mono 2.29x -3 g m -2. Then from the intercept of the curve we calculate B.4 m 3 g - In order to calculate ΔG we need to convert B s units to L/mol. We do that by multiplying by the molar weight of fibrinogen. B.4 m 3 g - x 34 g mol - x L m x 7 L mol -. Now the ΔG value is: B c solvent ΔG exp RT.394x ΔG exp 8.34x33 Δ G -5.6 kj/mol

7 (b) In order to calculate the orientation (angle Φ) of fibrinogen molecules orbed on the surface, we need to calculate length l at % coverage. From the fit obtained in part (a) Γ mono 2.29x -3 g m -2. Using the relationship: Γ mono Mwt # molecules N A 2.29x 34 3 # molecules x # molecules 4.56 x 5 molecules/m 2 The number of molecules per square meter (n total ) is defined as the no. of molecules in the y direction (n y ) multiplied by the no. of molecules in the x direction (n x ), where: molecule m n y molecules / m m 5 ntotal nx molecules / m 8 n.538 y m m nm 37.9nm cosφ nm Φ 37

8 (c) Using a first order kinetic orption model to describe the orption of fibrinogen on the solid surface at constant temperature, we get: dθ k c fib ( θ) kdes dt θ Integrating the above expression and solving for θ, we get the following: We also know that θ(t) A/A max, where, kc fib θ ( t) fib des t k c + k fib des A max Γ [ exp[ ( k c + k ) ]] Γ max.69 mg m -2, which can be read from table at c fib. g L -. Therefore A max.74 cm -. Using Excel or any solver program, we can minimize the sum of error between the experimental data in table 2 and the data obtained from the model. Therefore we find: k.98 L g - min x 5 M - min - k des 3.63 x -2 min - max k B k des 26.96L / g 34g / mol L / mol ΔG -5.5 kj/mol The values of ΔG obtained in parts (a) and (c) are very close to each other. The discrepancy between the two values is due to experimental error. Under the conditions applied, the equilibrium coverage reached is ca. 74%. This maximum coverage is achieved at the time when the coverage becomes independent of time. This can be observed from the following curve where the surface coverage becomes flat at ca. 6 min.

9 .8.7 Aexp and Amodel /cm Exp data. model time /min