Tempkin Isotherm Essay

Improved Essays
Tempkin isotherm contains a factor that explicitly takes into account adsorbing species-adsorbate interactions . This isotherm assumes that the heat of adsorption of all molecules in the layer decreases linearly with coverage due to adsorbate-adsorbate interaction and adsorption was characterized by a uniform distribution of binding energies, up to some maximum binding energy. Tempkin isotherm has generally been used in the linearized and rearranged form as following: Q_e⁡〖=β lnK_T+ β ln⁡〖C_e 〗 〗.............................6 where, KT is an equilibrium constant of binding corresponding to the maximum energy of binding (mg L-1) and the β is related to the heat of adsorption. Figure 9. shows a plot of Qe versus …show more content…
Lagergren’s first-order kinetic model
The pseudo-first-order kinetic model of Lagergren is more suitable for lower concentration of solute and its linear form is: log⁡〖(Q_e-Q_t )=log⁡〖Q_e- k_1/2.303 t〗 〗 .................................7
Where, Qt (mg g-1) is the amount of adsorbate adsorbed at time t (min); Qe (mg g-1) is the adsorption capacity in the equilibrium; k1 (min-1) is the rate constant of pseudo-first-order
…show more content…
The values of k2 for removal of iron by Tribulus Terrestris adsorbents was calculated from the slopes of the respective linear plots of t/Qt vs. t (Figure.10).The correlation coefficients, R2 was 0.994 for Tribulus Terrestris respectively suggest a strong relationship between the parameters and also explain that the process follows pseudo second order kinetics (Table 6 ).
Elovich kinetic model
Elovich model suggests that the chemisorptions, i.e. a chemical reaction, is probably the
Mechanism that controls the rate of adsorption. This model can be applied with success in liquid solution and the linear form of the Elovich equation is: Q_t= 1/β ln⁡αβ+ 1/β ln⁡t ..................................8
Where, α (mg g-1) is the initial sorption rate and β (g mg-1) is the desorption constant. The values of α and β can be calculated from the slope and intercept of the plot of Qt versus ln t

Related Documents

  • Decent Essays

    5. CONCLUSIONS Dry waste prawn shell converted to chitin and chitosan by using 3.5% of NaOH and 1N HCl with the temperature range from 20 oC-120 oC in the process of deprotinized, demineralized , decolourized and deacytilated .CMC was synthesized by carboxymethylation, as some of the –OH group of chitosan were substituted by –CH2COOH groups. Carboxymethyl chitosan was used as an adsorbent for the removal of copper from the wastewater. The removal efficiencies of copper were affected by different parameters such as pH, adsorbent dosage, temperature, contact time, initial metal ion concentration, anion and cation.…

    • 170 Words
    • 1 Pages
    Decent Essays
  • Decent Essays

    Calorimetry Lab Report

    • 684 Words
    • 3 Pages

    Max 629nm| ML of Solution|Absorbance|Conc. (mg/ml)||ML of Solution|Absorbance|Conc. (mg/ml)| 1.00|0.159|0.1848||1.00|0.003|0.1848| 2.00|0.303|0.3697||2.00|0.005|0.3697| 3.00|0.457|0.5545||3.00|0.007|0.5545| 4.00|0.616|0.7393||4.00|0.011|0.7393| 5.00|0.753|0.9242||5.00|0.014|0.9242| 6.00|0.948|1.109||6.00|0.018|1.109| Average concentration = 0.6469 mg/ml absorbance @ 503nm = 0.5393 absorbance @ 629nm = 0.0097 UNKNOWN II||L. MAX 503NM||||L. MAX 629 nm| ML OF SOLUTION|ABSORBANCE|CONC. (mg/ml)||ML OF SOLUTION|ABSORBANCE|CONC.…

    • 684 Words
    • 3 Pages
    Decent Essays
  • Improved Essays

    Uncatalyzed Reaction Lab

    • 882 Words
    • 4 Pages

    The slope of the graphs would increase by a factor of 2.76 because the concentration of KI would be doubled, a lot like the difference between trial 1 and trial 3. The rate for each trial would be 2.76 times larger, because the concentration of the reactants would be different. This would not change k for the reaction, it just changes the concentration with respect to volume. rate =…

    • 882 Words
    • 4 Pages
    Improved Essays
  • Decent Essays

    Kinetic Reaction Lab

    • 527 Words
    • 3 Pages

    K is the rate constant and t is the change in time. reaction rate=-1/a d[A]/dt=-1/b b[B]/dt=1/c d[C]/dt=1/d d[D]/dt Rate = K[A]x[B]y…

    • 527 Words
    • 3 Pages
    Decent Essays
  • Improved Essays

    Kcat Lab

    • 779 Words
    • 4 Pages

    It determines the limitation of the reaction. If Kcat/Km is large, either the Kcat is large or the Km is small. Within the reaction, this means that the turnover rate is relatively rapid and the concentration of substrate required to obtain half the maximum rate is small. (Oregon 2015) The reaction is controlled by the 2 concentration of substrate.…

    • 779 Words
    • 4 Pages
    Improved Essays
  • Improved Essays

    The position of the equilibrium depends on the two reactions. The percent yield of the reaction was obtained from the following equation: Percent Yield=amount of product (obtained) (g) X 100 amount of product (theoretical) (g) Procedure Results Table 1. Results of DigiMelt Melting point Range 98.7°C-100.3°C 1.6°C Table 2. Percent Yield of the reaction Initial weight 516mg Amount recovered 27mg Percent yield 4% Discussion…

    • 450 Words
    • 2 Pages
    Improved Essays
  • Improved Essays

    Beer's Law Lab

    • 944 Words
    • 4 Pages

    Many concepts and theories were covered throughout this lab. To find the equilibrium constant of FeSCN2+ one needed to find the equilibrium concentration of the compound. In order to do this, Beer’s Law, which relates absorption to concentration, was implemented. The reference solutions were created as a way to formulate a Beer’s Law plot by way of measuring a known concentration of FeSCN2+ and measuring its absorption; varying amounts of each reaction were mixed together in a solution to have multiple points on the plot. When mixing the compounds together, a dilution occurs, where both reactants have a…

    • 944 Words
    • 4 Pages
    Improved Essays
  • Decent Essays

    This is when the pack gets hotter. Exothermic is the release of heat. This is when the pack gets colder. The enthalpy of dissolution for these salts determines rather they can be used in the hot or cold packs. The enthalpy of dissolution is the measurement of energy released or absorbed when a solute dissolves in a solvent.…

    • 303 Words
    • 2 Pages
    Decent Essays
  • Improved Essays

    Exothermic Reaction Lab

    • 1985 Words
    • 8 Pages

    This lab was conducted to find the key differences between the quantity of heat (q) and ∆H of a substance. Different masses of substances CaCl2 and NH4NO3 were dissolved in water to see how changing mass impacts the q and ∆H of a substance. When the two substances dissolve, they become ions. CaCl2 dissolves in a reaction of: CaCl2 Ca2+aq + 2Cl1-aq while NH4NO3 dissolve in a reaction of: NH4NO3 NH41+aq + NO31-aq . As shown in these reactions, when ionic compounds dissolve in water, they break into their individual charged ions.…

    • 1985 Words
    • 8 Pages
    Improved Essays
  • Great Essays

    Part B: Change the amount of enzyme Table 4: The absorbances of 5 cuvettes that measured at 475nm during a given time interval using spectrophotometer. Time (s) Absorbance @ 475nm Cuvette 1 Cuvette 2 Cuvette 3 Cuvette 4 Cuvette 5 0 0.003 0.003 0.004 0.002 0.009 30 0.003…

    • 850 Words
    • 4 Pages
    Great Essays
  • Improved Essays

    The determination of the pH of a buffer solution and the pKa for of a weak acid Introduction A practical was carried out to show how the Henderson-Hasselbalch equation works and to apply and hone the skills of pipetting, buffer-making, pH-meter use and graphing. The primary goal of the practical was to determine the pKa value of a weak acid. The Ka is an acid dissociation constant, it is a quantitative measure of the strength of an acid in solution.…

    • 1043 Words
    • 4 Pages
    Improved Essays
  • Superior Essays

    The experiment that I will be conducting focuses on how fast one tablet of Alka seltzer can dissolve in various temperatures of water. The core of this procedure is to determine the impact that temperature has on how fast something dissolves. My hypothesis for this experiment would be that the Alka Seltzer that dissolves the fastest would be the one that was placed in the water with the highest temperature. There will be controlled and manipulated variables explained throughout the experiment. The two main things used in this experiment will be the solute and the solvent, which is Alka Seltzer and water.…

    • 1462 Words
    • 6 Pages
    Superior Essays
  • Improved Essays

    In this lab we learned the basics of enzymes and how they work. We were able to perform a quantitative assay of the activity of an enzyme in a tissue extract using a spectrophotometer. Also, we had to organize the data that was provided into tables and graphs so we could have the ability to test a few hypotheses. For example, we tested the whether the rate of the reaction was influenced by enzyme concentration, whether the activity of the enzyme was influenced by temperature and also too see if the activity of the enzyme was influenced by the pH of the solution, which you will then see the results on the graphs and charts that I provided. In figure one we show results of the trial run from the enzyme activity where the absorbance was at 500nm.…

    • 718 Words
    • 3 Pages
    Improved Essays
  • Improved Essays

    Abstract: In this lab our main objective is to find the empirical formula of MgO, magnesium oxide. To do this first we have to make sure when we burn Mg in the crucible and it reacts to with O. This lab experiment Mg is complicated by another factor. Mg is so reactive that it reacts with N in the air so some of the Mg which is suppose to react with O will react will N instead.…

    • 1338 Words
    • 6 Pages
    Improved Essays
  • Improved Essays

    Essay On Thermoregulation

    • 737 Words
    • 3 Pages

    Thermoregulation is the process of regulating the internal body temperature of animals within a certain range. Endotherms, such as mammals and birds, thermoregulate internally by generating heat from metabolic processes and are able to keep their body temperature high and relatively constant, adapting to the environment to maintain homeostasis. This type of thermoregulation requires a high metabolic rate, that enables organisms to be more physically active without having to rely on the environment, hence increases evolutionary fitness. Ectotherms, such as reptiles and fish obtain heat from the environment and hence have a slower metabolism and can tolerate fluctuations in their internal temperature. Ectotherms thermoregulate by adjusting their behaviour, for example, laying in the sun or submerging themselves in water (Mrowka & Reuter 2016; Griebeler 2013; Grady et al. 2014;…

    • 737 Words
    • 3 Pages
    Improved Essays