The Mechanisms of Adsorption - Assignment Example

Name Course Lecturer Date Introduction The application of activated carbon can be dated back to several hundreds of years ago. Importantly, activated carbon has been used over this time for such things as water purification and sugar syrup refinery, in which case it produces white sugar. The production of activated carbon occurs through heating of charcoal prepared from a range of carbon sources like wood, bone, or coconut. These sources of carbon are heated at very high temperatures in a stream of dry air or in a vacuum. The characteristic of activated carbon is that it has high porosity as well as high internal surface area that ranges from 500 to 1500 m2/gram of carbon. Because of this high porosity, activated carbon serves as an effective organic chemicals’ adsorber. Apart from activated carbon, we have other examples of adsorbents, which include alumina (Al2O3), silica gel (SiO2), molecular sieves as well as zeolites. The aim of this experiment therefore is to investigate the adsorption of molecules of acetic acid from the particular solution they are found in to the molecules found at the surface of the solid than to the bulk solution. Generally, one utilizes adsorbents whose particle size is small and particles whose surface have some imperfections like holes and cracks. These imperfections plays a key role of increasing the surface area for each unit mass. It increases greatly beyond the apparent geometrical area. The kind of interaction existing between the solid surface and the adsorbed molecule varies over a broad range from strong chemical bonding to weak van der Waal’s forces, which are non polar. There is an example of adsorption in which covalent and ionic bonding takes place. Consequently, this is the adsorption of such ions as chloride on silver chloride. In this case, silver chloride is ionic. On the other hand, an example covalent bonding is the adsorption of oxygen gas on different metals. In these two cases, the adsorption process is called chemisorptions. High heats of adsorption characterize this process. It is also very specific in nature since it relies on the chemical features or characteristics of both the adsorbed molecules and the surface molecules. Adsorption that comes from dipole forces or weaker van der Waal’s forces on the other hand is not specific in nature and it can occur in any system at moderate or low temperatures. In this case, such type of adsorption is called physical adsorption. Low heats of adsorption characterize it. The forces of physical adsorption are the same as those causing gases to condense into liquids and liquids to condense into solids. As the adsorbing molecule approaches the solid’s surface, the molecule makes some interaction with the molecule on the surface. This interaction tends to bring together and concentrate the adsorbing molecule on the solid’s adsorbing surface. Isotherms refer to a measure of activated carbon’s ability to adsorb organic materials. This measure can be obtained by use of an adsorption isotherm. The amount of organic materials adsorbed for every gram of solid relies on the particular area of the solid, the nature and temperature if the of the molecules involved, the equilibrium of the concentration of solute. From measurements made at constant pressure, one can come up with a plot of N versus c. in this case; N refers to the number of molecules adsorbed for every gram of solid. C on the other hand refers to the equilibrium of solute concentration. The name given to this equilibrium is adsorption isotherm. A lot of effort has been used to develop a theory of adsorption. This theory could explain the observations made on experimental facts. A theory formulated by Langmuir can be used to explain some simple systems. However, this theory is limited to cases that have only one layer of molecules being adsorbed at the surface. On the other hand, physical adsorption, for instance from vapor phase involves the formation of several adsorbed layers at greater pressures. However, this is different as far as chemisorptions from the gas phase or monolayer adsorption is concerned. Consequently, monolayer adsorption is differentiated by the fact that the quantity of adsorbed goes to a maximum value when the moderate concentrations have been used. The derivation of isotherm can be done from either equilibrium or kinetic arguments, despite the fact that it is mainly used as far as the chemisorptions of gases is concerned. There is a formula form appropriate to adsorption from solution is θ =kadsc/1+kadsc In this case, θ is the fraction of solid surface that the adsorbed molecules cover and Kads refers to the adsorption constant. Therefore, θ =N/Nm1. In this case, N refers to the number of moles adsorbed for every gram needed to form a monolayer. If this substitution is made and the equation rearranged, the formula that result is c/N=c/Nm+1/kadsNm. In case the Langmuir isotherm is enough description of the process of adsorption, then a graph of C/N against c will produce a straight line whose slope is 1/Nm. In case the area covered by an adsorbed molecule on the solid’s surface is known, the particular area, in square meters per gram is illustrated by the equation A=NmNoσ X 10-20 In this case, No stands for Avogadro’s number whereas σ is given in square angstroms Freundlich is an alternative isotherm equation The equation is N=Kca. In this case, K and a constants have no much physical significance although they can be computed by plotting o the log N against log c. Procedure 1. A flask was placed on a top lading balance protected from air movement. It was tare or set to zero. Activated carbon was added directly to the flask until the mass of activated carbon is about one gram. The exact mass was recorded to the nearest milligram. The activated carbon was weighed similarly into a further 6 flasks. 2. To every flask, 20 ml of water and 25 ml of acetic acid solution was added. The table below gives the suggested initial concentrations. These concentrations can be made through the dilution of about 0.5 M acid with water Required concentration, M (Nominal) Volume of 0.5 M acid in ml Volume of reagent water in ml 0 0 100 0.050 10 90 0.100 20 60 0.350 70 30 0.500 100 0 3. After tightly stoppering the samples, those that contained activated carbon were placed in a thermostat bath for 20 min at 250C. This was meant for equilibrium to be attained. During this time, the flasks were shaken every few minutes to help in achieving equilibrium. 4. To a flask that contained no activated carbon, 100ml of acid with 0.50 M was added. The sample was filtered accordingly. The sample was titrated in duplicate with standardized sodium hydroxide solution. Thymol blue was used as the indicator. This titration determines the actual acetic acid. 5. All samples were filtered into 250 ml conical flask using vacuum. The conical falsk was fitted with filters. 6. 0.25 M sodium hydroxide solution, which was standardized was titrated into each of the Two 20 ml aliquots samples. Results Volume of 0.5 M acid, mL Volume of reagent water Ml Readings 1 2 36.7 37.1 0 100 0.2 0.1 10 90 3.1 2.8 20 60 6.5 6.5 70 30 24.0 24.2 100 0 35.9 35.8 Weight of activated carbon Flask Weight 1 1.001 2 1.000 3 1.014 4 1.003 5 1.003 Report The actual concentration of acetic acid for every sample was calculated Volume of 0.5 M acid in ml Volume of reagent water in ml Concentration of acetic acid 0 100 0 10 90 0.005 M 20 60 0.01 M 70 30 0.035 M 100 0 0.05M If 0.5M=1L Therefore 0.5 M=1000 mL 10ml=10/1000*0.5=0.005M 20Ml=20/1000*0.5=0.01M 70ML=70/1000*0.5=0.035 100ML=100/1000*0.5=0.05 The number of moles present after and before adsorption was calculated using the final and initial concentrations of acetic acid in 100 mL of solution Before adsorption, the number of moles of acetic acid present in 100mL of solution is equivalent to 100/1000*0.5=0.05 moles After the adsorption the number of moles present = HC2H3O2(aq) + NaOH(s) ------> H2O(l) + NaC2H3O2(s) In this case, 38.85 of the 0.25 moles per liter of NaOH neutralized the acetic acid This means that if 0.25 moles=1000ml 38.85 ml=?moles =38.85/1000*0.25=0.01 moles Since the mole ratio is the same, it means the number of moles present in the final concentration is 0.01 moles N was computed. This represented the number of moles of acid adsorbed for every gram of activated carbon Flask Weight Moles N 1 1.001 0 0 2 1.000 0.005 M 0.00008 3 1.014 0.01 M 0.0002 4 1.003 0.035 M 0.0006 5 1.003 0.05M 0.0008 1 mole of HC2H3O2=1+24+3+32 g N=number of moles of acid adsorbed for every gram of carbon 60g=0.005M 1.000g=0.00008 An isotherm of N versus c was plotted N Concentration of acetic acid (c) 0 0 0.00008 0.005 M 0.0002 0.01 M 0.0006 0.035 M 0.0008 0.05M c/N against c was also plotted Nm was calculated from the slope By use of y-intercept and Nm, Kads was calculated Questions Some interactions of natural materials that demonstrate key surface reactions like adsorption and the mechanism of adsorption If a gas like SO2, NH3, CL2, CO, H2 and O2 is put in a closed container with charcoal, the gas pressure in the enclosed container decreases as the gas molecules concentrate on the charcoal’s surface. An organic dye like methylene blue turns colourless when mixed with charcoal as the molecules of the organic dye accumulate on the charcoal’s surface. Passing an aqueous raw sugar solution over beds of animal charcoal makes the solution to be colorless since charcoal adsorbs the coloring substance. In presence of silica gel, the air becomes dry since the gel adsorbs water molecules on the surface. Q 2. Relationship between charcoal and activated carbon. The thing that defines each of these materials Activated carbon has been processed such that it becomes extremely porous with large surface area for chemical reactions and adsorption. Q 3 Nature of difference existing between pesticide adsorption on clay and ion exchange at clay surfaces In pesticide adsorption, it is the high surface area of the clay particles caused by their small particles sizes as well as the smectites that does the adsorption. Smectites have expandable structures or swellings that ensure that the interlamellar surface is accessible to cations interchange with pesticides. In ion exchange, the surface area does not play a lot of role. It is more of chemical reaction. Q 4. Functions and differences between blank sample and control sample Control sample is found in any kind of experiment. It is not exposed to conditions. It helps in showing the changes that occur in a reaction. Blank sample on the other hand is found mainly in Spectrophotometry. It helps in the measurement of absorbance when there is no sample. Read More

From measurements made at constant pressure, one can come up with a plot of N versus c. in this case; N refers to the number of molecules adsorbed for every gram of solid. C on the other hand refers to the equilibrium of solute concentration. The name given to this equilibrium is adsorption isotherm. A lot of effort has been used to develop a theory of adsorption. This theory could explain the observations made on experimental facts. A theory formulated by Langmuir can be used to explain some simple systems.

However, this theory is limited to cases that have only one layer of molecules being adsorbed at the surface. On the other hand, physical adsorption, for instance from vapor phase involves the formation of several adsorbed layers at greater pressures. However, this is different as far as chemisorptions from the gas phase or monolayer adsorption is concerned. Consequently, monolayer adsorption is differentiated by the fact that the quantity of adsorbed goes to a maximum value when the moderate concentrations have been used.

The derivation of isotherm can be done from either equilibrium or kinetic arguments, despite the fact that it is mainly used as far as the chemisorptions of gases is concerned. There is a formula form appropriate to adsorption from solution is θ =kadsc/1+kadsc In this case, θ is the fraction of solid surface that the adsorbed molecules cover and Kads refers to the adsorption constant. Therefore, θ =N/Nm1. In this case, N refers to the number of moles adsorbed for every gram needed to form a monolayer.

If this substitution is made and the equation rearranged, the formula that result is c/N=c/Nm+1/kadsNm. In case the Langmuir isotherm is enough description of the process of adsorption, then a graph of C/N against c will produce a straight line whose slope is 1/Nm. In case the area covered by an adsorbed molecule on the solid’s surface is known, the particular area, in square meters per gram is illustrated by the equation A=NmNoσ X 10-20 In this case, No stands for Avogadro’s number whereas σ is given in square angstroms Freundlich is an alternative isotherm equation The equation is N=Kca.

In this case, K and a constants have no much physical significance although they can be computed by plotting o the log N against log c. Procedure 1. A flask was placed on a top lading balance protected from air movement. It was tare or set to zero. Activated carbon was added directly to the flask until the mass of activated carbon is about one gram. The exact mass was recorded to the nearest milligram. The activated carbon was weighed similarly into a further 6 flasks. 2. To every flask, 20 ml of water and 25 ml of acetic acid solution was added.

The table below gives the suggested initial concentrations. These concentrations can be made through the dilution of about 0.5 M acid with water Required concentration, M (Nominal) Volume of 0.5 M acid in ml Volume of reagent water in ml 0 0 100 0.050 10 90 0.100 20 60 0.350 70 30 0.500 100 0 3. After tightly stoppering the samples, those that contained activated carbon were placed in a thermostat bath for 20 min at 250C. This was meant for equilibrium to be attained. During this time, the flasks were shaken every few minutes to help in achieving equilibrium. 4. To a flask that contained no activated carbon, 100ml of acid with 0.

50 M was added. The sample was filtered accordingly. The sample was titrated in duplicate with standardized sodium hydroxide solution. Thymol blue was used as the indicator. This titration determines the actual acetic acid. 5. All samples were filtered into 250 ml conical flask using vacuum. The conical falsk was fitted with filters. 6. 0.25 M sodium hydroxide solution, which was standardized was titrated into each of the Two 20 ml aliquots samples. Results Volume of 0.5 M acid, mL Volume of reagent water Ml Readings 1 2 36.

7 37.1 0 100 0.2 0.1 10 90 3.

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