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Benzene and hexane are being considered as solvents to extract acetic acid from aqueous mixtures. At \(30^{\circ} \mathrm{C},\) distribution coefficients for the two solvents are \(\mathrm{K}_{\mathrm{B}}=0.098\) mass fraction acetic acid in benzene/mass fraction acetic acid in water and \(\mathrm{K}_{\mathrm{H}}=0.017\) mass fraction acetic acid in hexane/mass fraction acetic acid in water.(a) Based on the distribution coefficients only, which of the two solvents would you use and why? Demonstrate the logic of your decision by comparing the quantities of the two solvents required to reduce the acetic acid content in \(100 \mathrm{kg}\) of an aqueous solution from \(30 \mathrm{wt} \%\) to \(10 \mathrm{wt} \%\).(b) What other factors may be important in choosing between benzene and cyclohexane?

Short Answer

Expert verified
Based on distribution coefficients only, benzene would be a more suitable solvent due to a higher distribution coefficient (\(K_B\)) and lower quantity required for the extraction. However, there could be other factors like cost, safety, and equipment compatibility that might affect the final decision.

Step by step solution

01

Understanding the distribution coefficient

Distribution coefficients measure the distribution of a solute between two immiscible solvents. It is defined as the ratio of concentrations of a solute in the two solvents. In this particular exercise, for benzene \(K_B=0.098\) and for hexane \(K_H=0.017\). Higher values of K indicate better extraction efficiency.
02

Determining the more efficient solvent

Given that \(K_B> K_H\), it can be determined that benzene is more efficient at extracting acetic acid from aqueous mixtures than hexane.
03

Calculating the quantities of solvents

This step involves a bit of calculations. First, calculate the total weight of acetic acid in the solution. It was 30wt% of 100 kg initially, so its initial weight is \(30kg\). It needs to be reduced to 10wt%, which will be \(10kg\). Now, the weight of acetic acid to be extracted is \(30kg - 10kg = 20kg\). The quantities of solvents can be calculated using the given distribution coefficients. For solvent B, it would be \(20kg / 0.098 = 204.08kg\). For solvent H, it would be \(20kg / 0.017 = 1176.47kg\).
04

Making the decision

It is evident from the calculations that benzene requires less quantity (204.08 kg) than hexane (1176.47 kg) to reduce the acetic acid content from 30 wt% to 10wt%. Hence, benzene is the more suitable solvent based on the distribution coefficients.
05

Discussing other factors

While the distribution coefficient is the key factor, other factors may also be considered. These can include: cost of the solvent, available equipment and their compatibility with the solvent, toxicity and safety concerns, solvent recovery and disposal methods, and possible reactions between the solvent and the solute.

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Key Concepts

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

Chemical Engineering Principles
When considering the principles of chemical engineering, particularly in the context of separations like solvent extraction, it's crucial to remember that the process aims to separate components based on their differing physical properties. In the exercise at hand, the distribution coefficient is central to understanding how a solute, acetic acid, will partition itself between two phases: the aqueous phase and the organic solvent phase.

Solvent extraction relies on the concept of equilibrium between the two immiscible solvents, in this case, water, and the organic solvent (either benzene or hexane). To ensure an efficient separation process, chemical engineers must understand the solute's affinity for each phase, which is dictated by intermolecular forces, solubility, and the overall thermodynamics of the system. The role of the distribution coefficient, therefore, is to quantitatively describe this affinity and guide engineers in selecting the appropriate extraction conditions and solvents.
Solute Distribution
The distribution of a solute between two immiscible solvents is a fundamental concept in solvent extraction processes. The distribution coefficient, often denoted as K, serves as a measure of the preference or tendency of the solute to distribute itself between two distinct phases. It is essential to understand that solute distribution is affected by factors such as temperature, pH, and the nature of both the solute and solvents involved.

In the given problem, acetic acid distribution between aqueous mixtures and either benzene or hexane is represented by their respective distribution coefficients, with the larger coefficient indicating a greater affinity of the solute for the organic phase. This property is not constant and can be tailored to optimize separation by varying conditions such as pH or by adding modifying agents that alter the solute's chemical environment.
Solvent Extraction Efficiency
Efficiency in solvent extraction is a measure of how well a solvent can separate a desired solute from a mixture. More efficient solvents require less solvent to extract the same amount of solute. In the context of our exercise, benzene is the more efficient solvent over hexane for extracting acetic acid from an aqueous solution due to its higher distribution coefficient. However, efficiency isn't solely about how much solute can be extracted; considerations must also be made regarding how easily the solvent can be recovered and reused, the cost of the solvent, and the environmental and safety impacts of its use in industrial processes.
Extraction Process Calculations
Calculating the quantities of solvent needed for extraction is essential in designing and scaling up processes. The exercise demonstrates this by comparing the quantities of benzene and hexane required to decrease the concentration of acetic acid in water. By understanding the initial and desired concentration of the solute, along with the distribution coefficient, we can solve for the amount of solvent needed.

The real-world applicability of such calculations extends far beyond the classroom. In industrial processes, precise calculations define the economic viability and sustainability of the extraction operations. While the mathematics may appear simple, it encapsulates critical decisions regarding throughput, costs, and environmental considerations that determine the process's feasibility and optimization.

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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.

Air containing 20.0 mole \(\%\) water vapor at an initial pressure of 1 atm absolute is cooled in a 1 -liter sealed vessel from \(200^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\).(a) What is the pressure in the vessel at the end of the process? (Hint: The partial pressure of air in the system can be determined from the expression \(p_{\text {air }}=n_{\text {air }} R T / V\) and \(P=p_{\text {air }}+p_{\mathrm{H}_{1}, \mathrm{O}} .\) You may neglect the volume of the liquid water condensed, but you must show that condensation occurs.) (b) What is the mole fraction of water in the gas phase at the end of the process?(c) How much water (grams) condenses?

The vapor leaving the top of a distillation column goes to a condenser in which either total or partial condensation takes place. If a total condenser is used, a portion of the condensate is returned to the top of the column as \(r e f l u x\) and the remaining liquid is taken off as the overhead product (or distillate). (See Problem 6.63.) If a partial condenser is used, the liquid condensate is returned as reflux and the uncondensed vapor is taken off as the overhead product.The overhead product from an \(n\) -butane- \(n\) -pentane distillation column is 96 mole \(\%\) butane. The temperature of the cooling fluid limits the condenser temperature to \(40^{\circ} \mathrm{C}\) or higher.(a) Using Raoult's law, estimate the minimum pressure at which the condenser can operate as a partial condenser (i.e., at which it can produce liquid for reflux) and the minimum pressure at which it can operate as a total condenser. In terms of dew point and bubble point, what do each of these pressures represent for the given temperature?(b) Suppose the condenser operates as a total condenser at \(40^{\circ} \mathrm{C}\), the production rate of overhead product is \(75 \mathrm{kmol} / \mathrm{h}\), and the mole ratio of reflux to overhead product is \(1.5: 1 .\) Calculate the molar flow rates and compositions of the reflux stream and the vapor feed to the condenser.(c) Suppose now that a partial condenser is used, with the reflux and overhead product in equilibrium at \(40^{\circ} \mathrm{C}\) and the overhead product flow rate and reflux-to-overhead product ratio having the values given in Part (b). Calculate the operating pressure of the condenser and the compositions of the reflux and vapor feed to the condenser.

Dehydration of natural gas is necessary to prevent the formation of gas hydrates, which can plug valves and other components of a gas pipeline, and also to reduce potential corrosion problems. Water removal can be accomplished as shown in the following schematic diagram: Natural gas containing \(80 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O} / 10^{6} \mathrm{SCF}\) gas \(\left[\mathrm{SCF}=\mathrm{ft}^{3}(\mathrm{STP})\right]\) enters the bottom of an absorber at a rate of \(4.0 \times 10^{6}\) SCF/day. A liquid stream containing triethylene glycol (TEG, molecular weight \(=150.2\) ) and a small amount of water is fed to the top of the absorber. The absorber operates at 500 psia and \(90^{\circ} \mathrm{F}\). The dried gas leaving the absorber contains \(10 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O} / 10^{6} \mathrm{SCF}\) gas. The solvent leaving the absorber, which contains all the TEG-water mixture fed to the column plus all the water absorbed from the natural gas, goes to a distillation column. The overhead product stream from the distillation column contains only liquid water. The bottoms product stream, which contains TEG and water, is the stream recycled to the absorber.(a) Draw and completely label a flowchart of this process. Calculate the mass flow rate ( \(\left(\mathrm{b}_{\mathrm{m}} / \mathrm{day}\right)\) and volumetric flow rate (ft \(^{3}\) /day) of the overhead product from the distillation column. (b) The greatest possible amount of dehydration is achieved if the gas leaving the absorption column is in equilibrium with the solvent entering the column. If the Henry's law constant for water in TEG at \(90^{\circ} \mathrm{F}\) is \(0.398 \mathrm{psia} / \mathrm{mol}\) fraction, what is the maximum allowable mole fraction of water in the solvent fed to the absorber?(c) A column of infinite height would be required to achieve equilibrium between the gas and liquid at the top of the absorber. For the desired separation to be achieved in practice, the mole fraction of water in the entering solvent must be less than the value calculated in Part (b). Suppose it is \(80 \%\) of that value and the flow rate of TEG in the recirculating solvent is 37 Ib \(_{\mathrm{m}}\) TEG/lb \(_{\mathrm{m}}\) water absorbed in the column. Calculate the flow rate ( \(\left(\mathrm{b}_{\mathrm{m}} / \mathrm{day}\right)\) of the solvent stream entering the absorber and the mole fraction of water in the solvent stream leaving the absorber. (d) What is the purpose of the distillation column in the process? (Hint: Think about how the process would operate without it.)

The pressure in a vessel containing methane and water at \(70^{\circ} \mathrm{C}\) is 10 atm. At the given temperature, the Henry's law constant for methane is \(6.66 \times 10^{4}\) atm/mole fraction. Estimate the mole fraction of methane in the liquid.

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