/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 Various amounts of activated car... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Various amounts of activated carbon were added to a fixed amount of raw cane sugar solution \((48 \mathrm{wt} \%\) sucrose in water) at \(80^{\circ} \mathrm{C} .\) A colorimeter was used to measure the color of the solutions, \(R,\) which is proportional to the concentration of trace unknown impurities in the solution. The following data were obtained (adapted from the reference in Footnote \(20,\) p. 652 ):$$\begin{array}{|l|r|r|r|r|r|r|} \hline \text { kg carbon/kg dry sucrose } & 0 & 0.005 & 0.010 & 0.015 & 0.020 & 0.030 \\ \hline R \text { (color units/kg sucrose) } & 20.0 & 10.6 & 6.0 & 3.4 & 2.0 & 1.0 \\ \hline\end{array}.$$ The reduction in color units is a measure of the mass of impurities (the adsorbate) adsorbed on the carbon (the adsorbent).(a) The general form of the Freundlich isotherm is $$X_{i}^{*}=K_{\mathrm{F}} c_{i}^{\beta}$$ where \(X_{i}^{*}\) is the mass of \(i\) adsorbed/mass of adsorbent and \(c_{i}\) is the concentration of \(i\) in solution. Demonstrate that the Freundlich isotherm may be formulated for the system described above as $$\vartheta=K_{\mathrm{F}}^{\prime} R^{\beta}$$ where \(\vartheta\) is the \(\%\) removal of color/[mass of carbon/mass of dissolved sucrose]. Then determine \(K_{\mathrm{F}}^{\prime}\) and \(\beta\) by fitting this expression to the given data, using one of the methods in Section \(2.7 .\) (b) Calculate the amount of carbon that would have to be added to a vat containing \(1000 \mathrm{kg}\) of the 48 wt\% sugar solution at \(80^{\circ} \mathrm{C}\) for a reduction in color content to \(2.5 \%\) of the original value.

Short Answer

Expert verified
The solution involves deriving the Freundlich isotherm for the specific system, determining the Freundlich constants using the provided data and then applying these constants to calculate the required amount of activated carbon necessary to achieve specific removal of the color content. The numerical values for the constants and required activated carbon are dependent on the obtained plot or numerical analysis from the provided data.

Step by step solution

01

Derive the Freundlich Isotherm

First, it's important to establish the Freundlich isotherm specific for this system. This means relating \(X_{i}^{*}\), the mass of impurities adsorbed per mass of carbon, to \(c_{i}\), the concentration of impurities in the solution, using the Freundlich isotherm, \(X_{i}^{*}=K_{\mathrm{F}} c_{i}^{\beta}\). Here, the concentration of impurities is represented by \(R\), the color of the solution. Similarly, \(\%\) removal of color per [mass of carbon/mass of dissolved sucrose] is equal to the mass of \(i\) adsorbed per mass of carbon (\(X_{i}^{*}\)). Therefore, for this specific system, the Freundlich isotherm is: \(\vartheta=K_{\mathrm{F}}^{\prime} R^{\beta}\).
02

Determine the Freundlich constants

The Freundlich constants \(K_{\mathrm{F}}^{\prime}\) and \(\beta\), are obtained by fitting this derived isotherm to the provided data (kg carbon/kg dry sucrose and R), using logarithmic plotting or numerical analysis methods. This equation is in the form of a power function, so it can be converted to linear form by taking the logarithm of both sides: \(\log \vartheta=\log K_{\mathrm{F}}^{\prime} + \beta \log R\). This linear form can be plotted using the provided data, with the resulting slope equal to \(\beta\) and the intercept equal to \(\log K_{\mathrm{F}}^{\prime}\). Therefore, the constants \(K_{\mathrm{F}}^{\prime}\) and \(\beta\) can be determined using this line.
03

Calculate the required activated carbon

Upon finding the constants \(K_{\mathrm{F}}^{\prime}\) and \(\beta\), use them to calculate the required amount of activated carbon necessary to achieve a desired reduction in color content to \(2.5 \%\) of the original value. This is done using the derived Freundlich isotherm, solving for \(\vartheta\) (since it's equal to activated carbon required/kg dry sucrose) when \(R\) is \(2.5\%\) of the original value, and the mass of dry sucrose is known (in this case, considering the 48 wt% sugar solution).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Freundlich Isotherm
The Freundlich Isotherm is an empirical equation describing how solutes interact with surfaces in adsorption processes. It is particularly useful for heterogeneous surface energies, which occur in real-life scenarios. In our exercise, the isotherm takes the form \(X_{i}^{*}=K_{\mathrm{F}} c_{i}^{\beta}\), where \(X_{i}^{*}\) is the mass of the impurities adsorbed per mass of adsorbent (carbon), and \(c_{i}\) is the concentration in the solution.
By manipulating this, we get \(\vartheta=K_{\mathrm{F}}^{\prime} R^{\beta}\), which relates to the percent color removal per mass of carbon per mass of dissolved sucrose. Here, \(\vartheta\) represents a redefined term for our system.
This transformation allows us to link impurity concentration, indicated by color (\(R\)), to how much gets adsorbed on carbon. By fitting data to this model, constants \(K_{\mathrm{F}}^{\prime}\) and \(\beta\) can be identified, essential for predicting adsorption dynamics in various scenarios.
Adsorbent
An adsorbent is a material tasked with bonding to atoms, ions, or molecules (known collectively as adsorbates) during a process called adsorption. In this context, activated carbon plays the role of the adsorbent. It’s a highly porous substance with a large surface area, which makes it effective in capturing impurities.
The active sites on the carbon interact with various contaminants in the sugar solution, including trace impurities that impact color. Activated carbon's efficiency makes it extensively used in industries for water purification, air filtration, and sugar refining.
The amount of activated carbon utilized can be directly linked to how much of the impurities—or color—in the solution are removed. This relationship is critical when applying the Freundlich isotherm to understand the system's behavior.
Trace Impurities
In processes involving solutions like our sugar mixture, trace impurities are undesired tiny particles or substances that affect quality. Despite being in low concentrations, these impurities can significantly impact the overall appearance or properties of a solution.
Here, trace impurities affect the sugar solution’s color. The use of activated carbon serves to reduce these impurities by adsorbing them onto its surface, thereby enhancing the quality and clarity of the sugar solution.
Understanding and managing trace impurities is crucial in industries such as food processing, where the purity of ingredients is vital for product safety and consumer satisfaction.
Colorimetry
Colorimetry is a technique used to determine the concentration of colored compounds in solution by measuring the light absorption's intensity. In this exercise, colorimetry assists in quantifying trace impurities in the sugar solution by examining its color.
The measurement obtained, denoted as \(R\), reflects the amount of impurities because the more color present, the higher the concentration of these impurities. This measurement provides insights into how effective the adsorbent—activated carbon—has been in cleansing the solution.
Using colorimetry helps in accurately gauging the achievement of desired purity levels in solutions, making the process critical in quality control and assurance across many industries.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Penicillin is produced by fermentation and recovered from the resulting aqueous broth by extraction with butyl acetate. The penicillin distribution coefficient \(K\) (mass fraction of penicillin in the butyl acetate phase/mass fraction of penicillin in the water phase) depends strongly on the pH in the aqueous phase:$$\begin{array}{|r|c|c|c|}\hline \mathrm{pH} & 2.1 & 4.4 & 5.8 \\\\\hline K & 25.0 & 1.38 & 0.10 \\\\\hline\end{array}$$,This dependence provides the basis for the process to be described. Water and butyl acetate may be considered immiscible. The extraction is performed in the following three-unit process:\(\bullet\) After filtration, broth from a fermentor containing dissolved penicillin, other soluble impurities, and water is acidified in a mixing tank. The acidified broth, which contains 1.5 wt\% penicillin, is contacted with liquid butyl acetate in an extraction unit consisting of a mixer, in which the aqueous and organic phases are brought into intimate contact with each other, followed by a settling tank, in which the two phases separate under the influence of gravity. The pH of the aqueous phase in the extraction unit equals \(2.1 .\) In the mixer \(90 \%\) of the penicillin in the feed broth transfers from the aqueous phase to the organic phase.\(\bullet\) The two streams leaving the settler are in equilibrium with each other- -that is, the ratio of the penicillin mass fractions in the two phases equals the value of \(K\) corresponding to the pH of the aqueous phase \((=2.1 \text { in Unit } 1\) ). The impurities in the feed broth remain in the aqueous phase. The raffinate (by definition, the product stream containing the feed-solution solvent) leaving Extraction Unit 1 is sent elsewhere for further processing, and the organic extract (the product stream containing the extracting solvent) is sent to a second mixer-settler unit.\(\bullet\) In the second unit, the organic solution fed to the mixing stage is contacted with an alkaline aqueous solution that adjusts the pH of the aqueous phase in the unit to \(5.8 .\) In the mixer, \(90 \%\) of the penicillin entering in the organic feed solution transfers to the aqueous phase. Once again, the two streams emerging from the settler are in equilibrium. The aqueous extract is the process product.(a) Taking a basis of \(100 \mathrm{kg}\) of acidified broth fed to the first extraction unit, draw and completely label a flowchart of this process and carry out the degree-of-freedom analysis to show that all labeled variables can be determined. (Suggestion: Consider the combination of water, impurities, and acid as a single species and the alkaline solution as a second single species, since the components of these "pseudospecies" always stay together in the process.)(b) Calculate the ratios (kg butyl acetate required/kg acidified broth) and (kg alkaline solution required/kg acidified broth) and the mass fraction of penicillin in the product solution.(c) Briefly explain the following:(i) What is the likely reason for transferring most of the penicillin from an aqueous phase to an organic phase and then transferring most of it back to an aqueous phase, when each transfer leads to a loss of some of the drug? (ii) What is the purpose of acidifying the broth prior to the first extraction stage, and why is the extracting solution added to the second unit a base? (iii) Why are the two "raffinates" in the process the aqueous phase leaving the first unit and the organic phase leaving the second unit, and vice versa for the "extracts"? (Look again at the definitions of these terms.)(d) An alternative process for recovering the penicillin from the fermentation broth might involve evaporation to dryness. In that case, all the water simply is evaporated. Give two possible reasons for rejection of this alternative.

A fuel cell is an electrochemical device in which hydrogen reacts with oxygen to produce water and DC electricity. A 1-watt proton-exchange membrane fuel cell (PEMFC) could be used for portable applications such as cellular telephones, and a \(100-\mathrm{kW}\) PEMFC could be used to power an automobile. The following reactions occur inside the PEMFC:Anode: \(\quad \mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) Cathode: \(\quad \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}\) Overall: \(\quad \overline{\mathrm{H}}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}\) A flowchart of a single cell of a PEMFC is shown below. The complete cell would consist of a stack of such cells in series, such as the one shown in Problem 9.19.The cell consists of two gas channels separated by a membrane sandwiched between two flat carbonpaper electrodes- -the anode and the cathode- -that contain imbedded platinum particles. Hydrogen flows into the anode chamber and contacts the anode, where \(\mathrm{H}_{2}\) molecules are catalyzed by the platinum to dissociate and ionize to form hydrogen ions (protons) and electrons. The electrons are conducted throughthe carbon fibers of the anode to an extemal circuit, where they pass to the cathode of the next cell in the stack. The hydrogen ions permeate from the anode through the membrane to the cathode.Humid air is fed into the cathode chamber, and at the cathode \(\mathrm{O}_{2}\) molecules are catalytically split to form oxygen atoms, which combine with the hydrogen ions coming through the membrane and electrons coming from the external circuit to form water. The water desorbs into the cathode gas and is carried out of the cell. The membrane material is a hydrophilic polymer that absorbs water molecules and facilitates the transport of the hydrogen ions from the anode to the cathode. Electrons come from the anode of the cell at one end of the stack and flow through an extemal circuit to drive the device that the fuel cell is powering, while the electrons coming from the device flow back to the cathode at the opposite end of the stack to complete the circuit. is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. On the other hand, if there is not enough water in the cathode gas (less than \(85 \%\) relative humidity), the membrane dries out and cannot transport hydrogen efficiently, which also leads to reduced performance. 400-sell 300-yolt PEMFS anerates at stady state witha nonwer outnul of 36 k W, The air fod to It is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. On the other hand, if there is not enough water in the cathode gas (less than \(85 \%\) relative humidity), the membrane dries out and cannot transport hydrogen efficiently, which also leads to reduced performance.A 400-cell 300-volt PEMFC operates at steady state with a power output of 36 kW. The air fed to the cathode side is at \(20.0^{\circ} \mathrm{C}\) and roughly 1.0 atm (absolute) with a relative humidity of \(70.0 \%\) and a volumetric flow rate of \(4.00 \times 10^{3}\) SLPM (standard liters per minute). The gas exits at \(60^{\circ} \mathrm{C}\). (a) Explain in your own words what happens in a single cell of a PEMFC. (b) The stoichiometric hydrogen requirement for a PEMFC is given by \(\left(n_{\mathrm{Hz}}\right)_{\text {conanmad }}=I N / 2 F,\) where \(I\) is the current in amperes (coulomb/s), \(N\) is the number of single cells in the fuel cell stack, and \(F\) is the Faraday constant, 96,485 coulombs of charge per mol of electrons. Derive this expression. (Hint: Recall that since the cells are stacked in series the same current flows through each one, and the same quantity of hydrogen must be consumed in each single cell to produce that current at each anode.) (c) Use the expression of Part (b) to determine the molar rates of oxygen consumed and water generated in the unit with the given specifications, both in units of mol/min. (Remember that power = voltage \(\times\) current.) Then determine the relative humidity of the cathode exit stream, \(h_{\mathrm{r} \text { rout. }}\) (d) Determine the minimum cathode inlet flow rate in SLPM to prevent the fuel cell from flooding ( \(h_{\mathrm{r}, \text { out }}=100 \%\) ) and the maximum flow rate to prevent it from drying \(\left(h_{\mathrm{r}, \text { out }}=85 \%\right)\) .

In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20,000-liter cistem has been installed during construction of a new building. The cistem collects water from an HVAC (heating, ventilation, and air-conditioning) system designed to provide 2830 cubic meters of air per minute at \(22^{\circ} \mathrm{C}\) and \(50 \%\) relative humidity after converting it from ambient conditions \(\left(31^{\circ} \mathrm{C}, 70 \% \text { relative humidity }\right) .\) The collected condensate serves as the source of water for lawn maintenance. Estimate (a) the rate of intake of air at ambient conditions in cubic feet per minute and (b) the hours of operation required to fill the cistern.

A quantity of methyl acetate is placed in an open, transparent, three-liter flask and boiled long enough to purge all air from the vapor space. The flask is then sealed and allowed to equilibrate at \(30^{\circ} \mathrm{C},\) at which temperature methyl acetate has a vapor pressure of \(269 \mathrm{mm}\) Hg. Visual inspection shows \(10 \mathrm{mL}\) of liquid methyl acetate present.(a) What is the pressure in the flask at equilibrium? Explain your reasoning.(b) What is the total mass (grams) of methyl acetate in the flask? What fraction is in the vapor phase at equilibrium?(c) The above answers would be different if the species in the vessel were ethyl acetate because methyl acetate and ethyl acetate have different vapor pressures. Give a rationale for that difference.

An adult inhales approximately 12 times per minute, taking in about 500 mL of air with each inhalation. Oxygen and carbon dioxide are exchanged in the lungs, but there is essentially no exchange of nitrogen. The exhaled air has a mole fraction of nitrogen of 0.75 and is saturated with water vapor at body temperature, \(37^{\circ} \mathrm{C}\). If ambient conditions are \(25^{\circ} \mathrm{C}, 1\) atm, and \(50 \%\) relative humidity, what volume of liquid water (mL) would have to be consumed over a two-hour period to replace the water loss from breathing? How much would have to be consumed if the person is on an airplane where the temperature, pressure, and relative humidity are respectively \(25^{\circ} \mathrm{C}, 1 \mathrm{atm},\) and \(10 \% ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.