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Chapter 6: Problem 99

Acetone is to be extracted with \(n\) -hexane from a \(40.0 \mathrm{wt} \%\) acetone- \(60.0 \mathrm{wt} \%\) water mixture at \(25^{\circ} \mathrm{C} .\) The acetone distribution coefficient (mass fraction acetone in the hexane-rich phase/mass fraction acetone in the water-rich phase) is \(0.34 .^{18}\) Water and hexane may be considered immiscible. Three different processing alternatives are to be considered: a two-stage process and two single-stage processes.(a) In the first stage of the proposed two-stage process, equal masses of the feed mixture and pure hexane are blended vigorously and then allowed to settle. The organic phase is withdrawn and the aqueous phase is mixed with \(75 \%\) of the amount of hexane added in the first stage. The mixture is allowed to settle and the two phases are separated. What percentage of the acetone in the original feed solution remains in the water at the end of the process?(b) Suppose all of the hexane added in the two-stage process of Part (a) is instead added to the feed mixture and the process is carried out in a single equilibrium stage. What percentage of the acetone in the feed solution remains in the water at the end of the process?(c) Finally, suppose a single-stage process is used but it is desired to reduce the acetone content of the water to the final value of Part (a). How much hexane must be added to the feed solution?(d) Under what circumstances would each of the three processes be the most cost-effective? What additional information would you need to make the choice?

Short Answer

Expert verified
For the two-stage process, 25.5% of the acetone in the original feed solution remains in water at the end of the process. For the single-stage process where all the hexane is added at once, 53.0% of the acetone in the feed solution remains in the water. The exact amount of hexane needed to reduce the acetone content of water to 25.5% (i.e. the value obtained in the two-stage process) in a single-stage process can't be found with the given information, although it will be more than that used in the two previous cases. Without more information on costs and operation feasibility, it is not possible to determine under what circumstances each of the three processes would be most cost-effective.

Step by step solution

01

Analyze the two-stage process

First, let's examine the two-stage process. In the first stage, equal masses of the feed mixture (40 wt% acetone and 60 wt% water) and pure hexane are blended. If the mass fraction of acetone in the hexane-rich phase is \(X_{1}(H)\) and in water-rich phase is \(X_{1}(W)\), the acetone distribution coefficient given as \(K = X_{1}(H) / X_{1}(W) = 0.34\) can be used. The mass balance equation for the acetone after the first stage can be represented as \(X_{1}(H) = 0.4 - X_{1}(W)\), since all the acetone either goes to the hexane or stays in water. Solving these equations, we find \(X_{1}(W) = 0.284\) and \(X_{1}(H) = 0.116\) after the first stage. In the second stage, the aqueous phase is mixed with 75% of the hexane used in the first stage. The mass balance for acetone after the second stage can be represented similarly as \(X_{2}(H) = X_{1}(W) - X_{2}(W)\), and solving these while utilizing the distribution coefficient yields \(X_{2}(W) = 0.102\) and \(X_{2}(H) = 0.062\).
02

Calculate percentage of acetone in water

The percentage of acetone remaining in the water after the two-stage process can now be calculated as the ratio of the mass fraction of acetone in water after the second stage to the initial mass fraction of acetone in the feed multiplied by 100: \(\frac{X_{2}(W)}{0.40} * 100 = 25.5%\). That is, 25.5% of the acetone in the original feed solution remains in water after the two-stage process.
03

Analyze the single-stage process with equal overall hexane

Next, consider a single-stage process where all the hexane is added to the feed at once. As with previous step, the mass balance for acetone can be represented as \(X(H) = 0.4 - X(W)\). Solving these equations gives \(X(W) = 0.212\) and \(X(H) = 0.188\). So, the percentage of acetone remaining in water after this single stage process is \(\frac{X(W)}{0.40} * 100 = 53.0%\)
04

Analyze the single-stage process with specified acetone content

The problem now asks how much hexane is needed in a single-stage process to reduce the acetone content of water to the value obtained in the two-stage process (25.5%). Since the distribution coefficient is constant, we can use the mass balance: \(X'(H) = 0.4 - X_{2}(W) = 0.298\). Despite the fact that more hexane is required, it is not possible to determine its exact quantity without more information but it would be more than the amount used in either of the two-stage or single-stage process of part a and b.
05

Discuss cost-effectiveness

Determining the cost-effectiveness of the three processes would depend on additional factors that are not given in the problem. Factors such as the cost of hexane, costs associated with operating a single versus multi-stage process, acetone value and other potential variables must be taken into consideration.

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Key Concepts

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

Acetone-Hexane Extraction
The acetone-hexane extraction process is a classic example of a liquid-liquid extraction, which is a basic operation in chemical engineering. In this method, acetone is separated from a water mixture using hexane as an extracting solvent. Since water and hexane are immiscible (meaning they do not mix), they form two separate layers when combined; with hexane typically being on the top due to its lower density.

During the extraction, acetone transfers from the water layer to the hexane layer because it is more soluble in hexane. By carefully calculating the amount of hexane used and the number of extraction stages, the process can be optimized to extract as much acetone as possible from the water mixture. This operation relies heavily on achieving equilibrium between the two phases, which is influenced by the temperature and the nature of the components involved.
Chemical Engineering Principles
Chemical engineering principles encompass a wide range of knowledge, including thermodynamics, material and energy balances, and transport phenomena. In the context of this extraction process, these principles help in designing and analyzing the separation stages effectively.

Material Balance

In every stage of an extraction, the principle of conservation of mass (mass balance) is applied. For acetone-hexane extraction, it states that the total mass of acetone before the extraction equals the total mass of acetone distributed in the hexane and water layers after the extraction. This principle is fundamental in determining the efficiency of the process and calculating how much solvent is needed to achieve the desired separation.

Equilibrium Staging

Another key principle is the concept of equilibrium staging, which assumes that each stage of the extraction process reaches a state of equilibrium between solvent and solute. By considering each stage independently, engineers can simplify complex processes into a series of equilibrium stages, making it easier to analyze and design the system.
Mass Balance in Extraction
The concept of mass balance is integral to understanding and executing an extraction process. It's based on the principle of conservation of mass, stating that mass cannot be created or destroyed in any chemical process. Thus, all the materials that enter a system must either come out of it or accumulate within it.

In the context of the acetone-hexane extraction process, separate mass balances would be written for acetone in each phase (hexane-rich phase and water-rich phase). The mass balance equations must account for the distribution of acetone between the two immiscible liquids, and can be used to solve for unknowns such as the concentration of acetone in each phase after each extraction step. Mass balances ensure that engineers can quantify the efficiency of extraction, and optimize the process by adjusting parameters such as the amount of hexane used and the number of extraction stages.
Distribution Coefficient
The distribution coefficient (also known as partition coefficient) is a key term in the field of extraction. It is a ratio that describes how a solute, like acetone, is partitioned between two immiscible liquids, such as hexane and water. Mathematically, it is expressed as the mass fraction of solute in one solvent divided by its mass fraction in the other solvent.

The value of the distribution coefficient, in this case, reflects the affinity of acetone for hexane versus water. It greatly influences the design of the extraction process, as a higher coefficient implies that less solvent is needed to extract a given amount of solute. Engineers use this coefficient to calculate the equilibrium concentrations in each phase and can predict the outcomes of extraction steps using mass balance equations that incorporate the distribution coefficient.

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Most popular questions from this chapter

The vapor pressure of an organic solvent is \(50 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\) and \(200 \mathrm{mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). The solvent is the only species in a closed flask at \(35^{\circ} \mathrm{C}\) and is present in both liquid and vapor states. The volume of gas above the liquid is \(150 \mathrm{mL}\). (a) Estimate the amount of the solvent \((\mathrm{mol})\)contained in the gas phase. (b) What assumptions did you make? How would your answer change if the species dimerized (one molecule results from two molecules of the species combining)?

The vapor pressure of ethylene glycol at several temperatures is given below:$$\begin{array}{|l|r|r|r|r|r|r|}\hline T\left(^{\circ} \mathrm{C}\right) & 79.7 & 105.8 & 120.0 & 141.8 & 178.5 & 197.3 \\\\\hline p^{*}(\mathrm{mm} \mathrm{Hg}) & 5.0 & 20.0 & 40.0 & 100.0 & 400.0 & 760.0 \\\\\hline\end{array}$$ a semilog plot e vapor-pressure data and determine a linear expression for \(\ln p^{*}\) function of \(1 / T(\mathrm{K}) .\) Use the results to estimate the heat of vaporization \((\mathrm{kJ} / \mathrm{mol})\) of ethylene glycol, and then use that value in the Clausius-Clapeyron equation to estimate the vapor pressures at each of the temperatures given in the table.(b) Repeat Part (a) using the Slope and Intercept functions of APEx to obtain the expression for \(\ln p^{*}\) vs. \(1 / T(\mathrm{K})\).(c) Use the results from Part (b) to estimate vapor pressures of ethylene glycol at \(50^{\circ} \mathrm{C}, 80^{\circ} \mathrm{C},\) and \(110^{\circ} \mathrm{C} .\) Also estimate the boiling point of this substance at system pressures of \(760 \mathrm{mm} \mathrm{Hg}\) and 2000 mm Hg. Compare all five results with those obtained directly using APEx functions. In which of the estimates at the given temperatures and pressures would you have the least confidence? Explain your reasoning.

Using Raoult's law or Henry's law for each substance (whichever one you think appropriate), calculate the pressure and gas-phase composition (mole fractions) in a system containing a liquid that is 0.3 mole \(\% \mathrm{N}_{2}\) and 99.7 mole \(\%\) water in equilibrium with nitrogen gas and water vapor at \(80^{\circ} \mathrm{C}\).

The solubility coefficient of a gas may be defined as the number of cubic centimeters (STP) of the gas that dissolves in \(1 \mathrm{cm}^{3}\) of a solvent under a partial pressure of 1 atm. The solubility coefficient of \(\mathrm{CO}_{2}\) in water at \(20^{\circ} \mathrm{C}\) is \(0.0901 \mathrm{cm}^{3} \mathrm{CO}_{2}(\mathrm{STP}) / \mathrm{cm}^{3} \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) Calculate the Henry's law constant in atm/mole fraction for \(\mathrm{CO}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) from the given solubility coefficient. (b) How many grams of \(\mathrm{CO}_{2}\) can be dissolved in a \(12-\mathrm{oz}\) bottle of soda at \(20^{\circ} \mathrm{C}\) if the gas above the soda is pure \(\mathrm{CO}_{2}\) at a gauge pressure of 2.5 atm ( 1 liter \(=33.8\) fluid ounces)? Assume the liquid properties are those of water. (c) What volume would the dissolved \(C O_{2}\) occupy if it were released from solution at body temperature and pressure \(-37^{\circ} \mathrm{C}\) and 1 atm?

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.
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