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Chapter 4: Problem 73

Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right),\) an important solvent and intermediate in the production of many other chemicals, is produced by bubbling chlorine gas through liquid benzene in the presence of ferric chloride catalyst. In an undesired side reaction, the product is further chlorinated to dichlorobenzene, and in a third reaction the dichlorobenzene is chlorinated to trichlorobenzene. The feed to a chlorination reactor consists of essentially pure benzene and a technical grade of chlorine gas (98 wt\% \(\mathrm{Cl}_{2}\), the balance gaseous impurities with an average molecular weight of 25.0 ). The liquid output from the reactor contains \(65.0 \mathrm{wt} \% \mathrm{C}_{6} \mathrm{H}_{6}, 32.0 \% \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}, 2.5 \% \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) and \(0.5 \%\) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}_{3} .\) The gaseous output contains only \(\mathrm{HCl}\) and the impurities that entered with the chlorine. (a) You wish to determine (i) the percentage by which benzene is fed in excess, (ii) the fractional conversion of benzene, (iii) the fractional yield of monochlorobenzene, and (iv) the mass ratio of the gas feed to the liquid feed. Without doing any calculations, prove that you have enough information about the process to determine these quantities. (b) Perform the calculations. (c) Why would benzene be fed in excess and the fractional conversion kept low? (d) What might be done with the gaseous effluent? (e) It is possible to use 99.9\% pure ("reagent-grade") chlorine instead of the technical grade actually used in the process. Why is this probably not done? Under what conditions might extremely pure reactants be called for in a commercial process? (Hint: Think about possible problems associated with the impurities in technical grade chemicals.)

Short Answer

Expert verified
(a) We have enough information given in the problem to calculate all the requested quantities. (b) (i) The percentage by which benzene is fed in excess is 65%. (ii) The fractional conversion of benzene is 35%. (iii) The fractional yield of monochlorobenzene is 91%. (iv) The mass ratio of the gas feed to the liquid feed is 49:1. (c) Benzene is fed in excess to drive the reaction to completion and ensure a high yield of chlorobenzenes. The fractional conversion is kept low to limit the production of the unwanted byproducts. (d) The gaseous effluent might be used in other processes, or treated and disposed of accordingly. (e) Using reagent-grade chlorine instead of technical grade would increase the costs of the operation. It might be necessary to use reagent-grade chlorine if the impurities interfere with the reaction or the quality of the product.

Step by step solution

01

Validate the Information Present

All the information given in the problem is sufficient to determine the quantities asked. We know the input and output compositions. The balanced chemical reactions for the chlorination of benzene, dichlorobenzene and trichlorobenzene are given implicitly. Given the molecular weights, percentage weight of each species in feed and product and the concept of mass balance, we can find the excess benzene feed, conversions, yield of monochlorobenzene and the mass ratio of feed gases to liquid.
02

Calculations

(i) Using the given compositions, a material balance on \( \mathrm{C}_{6} \mathrm{H}_{6} \) would give the amount of unreacted benzene left:(0.65g benzene (in exit stream)) = (1g benzene (in feed))- (0.32g chlorobenzene + 0.025g dichlorobenzene + 0.005g trichlorobenane) produced in reactor. So, it is clear that we start off with an initial mass of 1g of benzene in feed. But, we have only reacted 0.35g of it and the remaining 0.650g of benzene is left unreacted. Hence, percentage by which benzene is fed in excess is given by: (0.650mg benzene left unreacted / 1g Benzene fed)*100 = 65% excess of Benzene fed. (ii) Fractional conversion of benzene is the reacted fraction out of the total fed: (1g Benzene - 0.650g)/1g Benzene = 0.35 or 35% is the fractional conversion of Benzene. (iii) Fractional yield of monochlorobenzene is the moles of Monochlorobenzene produced per mole of benzene reacted: (0.32g Monochlorobenzene produced / 0.35g Benzene reacted)= 0.91 or 91%. (iv) Let the feed rate of chlorine be A gram moles. Then, we have (0.98A/71) of \( \mathrm{Cl}_{2} \) in it and (0.02A/25) as impurities. A material balance on chlorine gives: (0.98A/71) = (0.32(1-x) + 2*0.025(1-y) + 3*0.005(1-z)) (as 1 mole of benzene reacted gives 1 mole of \( \mathrm{HCl} \)), where x,y and z are the fractions converted.Therefore, (A/71) = (0.32(1-x) + 2*0.025(1-y) + 3*0.005(1-z)) + ((0.32x + 0.025*2y + 0.005*3z))On solving, A = 71*(0.35 + 0.35) = 49g mol. Therefore the mass ratio of gas feed to liquid feed is 49:1.
03

Review of Industrial Implications

(c) Benzene may be fed in excess to drive the reaction towards completion and ensure a high yield of chlorobenzenes. Limiting the fractional conversion keeps the production of dichlorobenzene and trichlorobenzene low, which are undesired. (d) The gaseous effluent might be treated to remove HCl and impurities. It could be sent to a waste treatment plant or used in other processes that need HCl. (e) Using reagent-grade chlorine would increase the costs of the industrial process. It might only be necessary if the impurities in technical grade reactants interfered with the reaction or the purity of the desired product. Industrial processes usually tolerate a certain level of impurities.
04

Consideration of Purity

If the impurities start interfering with the quality of product or the operation of the equipment (catalyst deactivation, operational problems), it then makes sense to use a purer reactant. However, this would likely raise the operational costs so it would need to be a careful trade-off analysis between maintaining the equipment and the cost of the purer reactant.

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Key Concepts

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

Material Balances
Understanding material balances is crucial in chemical reaction engineering, particularly when designing and operating chemical reactors. The main goal is to account for all material entering and leaving a system. In chemical processes like chlorination, tracking the materials helps ensure efficiency and product quality.
In the chlorobenzene production process, we track the flow of benzene and chlorine through the reactor. We know the input composition (pure benzene and 98% pure chlorine gas) and the output composition (containing various chlorinated benzenes). To establish a material balance, we calculate the amount of each reactant entering the reactor and compare it to the products generated, assessing both unreacted and converted amounts.
Typically, we begin by establishing the basis (for example, assuming a certain weight of the benzene feed) and then calculate the conversion of benzene based on the products formed (chlorinated compounds). Excess feed of benzene may be calculated by comparing the unreacted and initial amounts, revealing how much of a reactant is left after the reaction takes place.
  • Ensures accountability of all materials in the process
  • Enables calculation of excess reactants, conversions, and yields
  • Aids in optimizing reactor design and operation
Reactions and Stoichiometry
Chemical reactions and stoichiometry are foundational principles in chemical engineering that dictate the proportions of reactants and products involved in a reaction. In the case of chlorobenzene production, we deal with multiple reactions where benzene is sequentially chlorinated.
The stoichiometry of each reaction indicates how many moles of benzene react to form chlorobenzene, dichlorobenzene, and trichlorobenzene. For each mole of benzene used, stoichiometric coefficients will inform how much chlorine is needed and how many moles of each product are generated.
In this process, the stoichiometry helps in balancing the reactions and assessing conversion and selectivity. For example, if one mole of benzene forms one mole of chlorobenzene, the fractional yield can be determined. This allows engineers to predict how changes in feed or conditions might affect the product distribution.
  • Helps calculate reactant needs and product formations
  • Supports conversion and yield assessments
  • Facilitates understanding of reaction pathways and selectivity
Industrial Chemical Processes
Industrial chemical processes require a consideration of economic viability and technical optimization. The production of chlorobenzene is a valuable process in the chemical industry, used as a solvent and intermediate for producing more complex chemicals.
In running these processes, engineers aim to optimize conditions to maximize desirable product yield and minimize waste and byproduct formation, such as dichlorobenzene and trichlorobenzene. An excess feed of benzene vs. fractional conversion is a typical strategy to control these unwanted by-products, as it provides a mechanism to steer the reaction towards the preferred monoproduct.
The use of technical-grade vs. reagent-grade chlorine reflects a balance between cost-efficiency and purity demands. While reagent-grade chemicals are more expensive, they are used only when impurities threaten the desired process outcomes or equipment integrity (e.g., causing catalyst deactivation).
  • Focuses on economic feasibility and production efficiency
  • Balances raw material costs and product purity requirements
  • Utilizes optimization techniques to enhance product yields

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Most popular questions from this chapter

The gas-phase reaction between methanol and acetic acid to form methyl acetate and water takes place in a batch reactor. When the reaction mixture comes to equilibrium, the mole fractions of the four reactive species are related by the reaction equilibrium constant $$K_{y}=\frac{y_{C} y_{D}}{y_{A} y_{B}}=4.87$$ (a) Suppose the feed to the reactor consists of \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0}, n_{\mathrm{D} 0},\) and \(n_{10}\) gram-moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and an inert gas, I, respectively. Let \(\xi\) be the extent of reaction. Write expressions for the gram-moles of each reactive species in the final product, \(n_{\mathrm{A}}(\xi), n_{\mathrm{B}}(\xi), n_{\mathrm{C}}(\xi),\) and \(n_{\mathrm{D}}(\xi) .\) Then use these expressions and the given equilibrium constant to derive an equation for \(\xi_{c}\), the equilibrium extent of reaction, in terms of \(\left.n_{\mathrm{A} 0}, \ldots, n_{10} . \text { (see Example } 4.6-2 .\right)\) (b) If the feed to the reactor contains equimolar quantities of methanol and acetic acid and no other species, calculate the equilibrium fractional conversion. (c) It is desired to produce 70 mol of methyl acetate starting with 75 mol of methanol. If the reaction proceeds to equilibrium, how much acetic acid must be fed? What is the composition of the final product? (d) Suppose it is important to reduce the concentration of methanol by making its conversion at equilibrium as high as possible, say 99\%. Again assuming the feed to the reactor contains only methanol and acetic acid and that it is desired to produce 70 mol of methyl acetate, determine the extent of reaction and quantities of methanol and acetic acid that must be fed to the reactor. (e) If you wanted to carry out the process of Part (b) or (c) commercially, what would you need to know besides the equilibrium composition to determine whether the process would be profitable? (List several things.)

A mixture of 75 mole \(\%\) methane and 25 mole \(\%\) hydrogen is burned with \(25 \%\) excess air. Fractional conversions of \(90 \%\) of the methane and \(85 \%\) of the hydrogen are achieved; of the methane that reacts, \(95 \%\) reacts to form \(\mathrm{CO}_{2}\) and the balance reacts to form CO. The hot combustion product gas passes through a boiler in which heat transferred from the gas converts boiler feedwater into steam. (a) Calculate the concentration of \(\mathrm{CO}\) (ppm) in the stack gas. (b) The CO in the stack gas is a pollutant. Its concentration can be decreased by increasing the percent excess air fed to the furnace. Think of at least two costs of doing so. (Hint: The heat released by the combustion goes into heating the combustion products; the higher the combustion product temperature, the more steam is produced.)

In an absorption tower (or absorber), a gas is contacted with a liquid under conditions such that one or more species in the gas dissolve in the liquid. A stripping tower (or stripper) also involves a gas contacting a liquid, but under conditions such that one or more components of the feed liquid come out of solution and exit in the gas leaving the tower. A process consisting of an absorption tower and a stripping tower is used to separate the components of a gas containing 30.0 mole \(\%\) carbon dioxide and the balance methane. A stream of this gas is fed to the bottom of the absorber. A liquid containing 0.500 mole\% dissolved \(\mathrm{CO}_{2}\) and the balance methanol is recycled from the bottom of the stripper and fed to the top of the absorber. The product gas leaving the top of the absorber contains 1.00 mole \(\% \mathrm{CO}_{2}\) and essentially all of the methane fed to the unit. The CO_-rich liquid solvent leaving the bottom of the absorber is fed to the top of the stripper and a stream of nitrogen gas is fed to the bottom. Ninety percent of the \(\mathrm{CO}_{2}\) in the liquid feed to the stripper comes out of solution in the column, and the nitrogen/CO_stream leaving the column passes out to the atmosphere through a stack. The liquid stream leaving the stripping tower is the \(0.500 \% \mathrm{CO}_{2}\) solution recycled to the absorber. The absorber operates at temperature \(T_{\mathrm{a}}\) and pressure \(P_{\mathrm{a}}\) and the stripper operates at \(T_{\mathrm{s}}\) and \(P_{\mathrm{s}}\) Methanol may be assumed to be nonvolatile- -that is, none enters the vapor phase in either column and \(\mathrm{N}_{2}\), may be assumed insoluble in methanol. (a) In your own words, explain the overall objective of this two-unit process and the functions of the absorber and stripper in the process. (b) The streams fed to the tops of each tower have something in common, as do the streams fed to the bottoms of each tower. What are these commonalities and what is the probable reason for them? (c) Taking a basis of 100 mol/h of gas fed to the absorber, draw and label a flowchart of the process. For the stripper outlet gas, label the component molar flow rates rather than the total flow rate and mole fractions. Do the degree-of-freedom analysis and write in order the equations you would solve to determine all unknown stream variables except the nitrogen flow rate entering and leaving the stripper. Circle the variable(s) for which you would solve each equation (or set of simultaneous equations), but don't do any of the calculations yet. (d) Calculate the fractional \(\mathrm{CO}_{2}\) removal in the absorber (moles absorbed/mole in gas feed) and the molar flow rate and composition of the liquid feed to the stripping tower. (e) Calculate the molar feed rate of gas to the absorber required to produce an absorber product gas flow rate of \(1000 \mathrm{kg} / \mathrm{h}\). (f) Would you guess that \(T_{\mathrm{s}}\) would be higher or lower than \(T_{\mathrm{a}} ?\) Explain. (Hint: Think about what happens when you heat a carbonated soft drink and what you want to happen in the stripper.) What about the relationship of \(P_{\mathrm{s}}\) to \(P_{\mathrm{a}} ?\) (g) What properties of methanol would you guess make it the solvent of choice for this process? (In more general terms, what would you look for when choosing a solvent for an absorption-stripping process to separate one gas from another?)

A drug (D) is produced in a three-stage extraction from the leaves of a tropical plant. About 1000 kg of leaf is required to produce 1 kg of the drug. The extraction solvent (S) is a mixture containing 16.5 wt\% ethanol (E) and the balance water (W). The following process is carried out to extract the drug and recover the solvent. 1\. A mixing tank is charged with \(3300 \mathrm{kg}\) of \(\mathrm{S}\) and \(620 \mathrm{kg}\) of leaf. The mixer contents are stirred for several hours, during which a portion of the drug contained in the leaf goes into solution. The contents of the mixer are then discharged through a filter. The liquid filtrate, which carries over roughly \(1 \%\) of the leaf fed to the mixer, is pumped to a holding tank, and the solid cake (spent leaf and entrained liquid) is sent to a second mixer. The entrained liquid has the same composition as the filtrate and a mass equal to \(15 \%\) of the mass of liquid charged to the mixer. The extracted drug has a negligible effect on the total mass and volume of the spent leaf and the filtrate. 2\. The second mixer is charged with the spent leaf from the first mixer and with the filtrate from the previous batch in the third mixer. The leaf is extracted for several more hours, and the contents of the mixer are then discharged to a second filter. The filtrate, which contains \(1 \%\) of the leaf fed to the second mixer, is pumped to the same holding tank that received the filtrate from the first mixer, and the solid cake- -spent leaf and entrained liquid - is sent to the third mixer. The entrained liquid mass is \(15 \%\) of the mass of liquid charged to the second mixer. 3\. The third mixer is charged with the spent leaf from the second mixer and with \(2720 \mathrm{kg}\) of solvent \(\mathrm{S}\). The mixer contents are filtered; the filtrate, which contains \(1 \%\) of the leaf fed to the third mixer, is recycled to the second mixer; and the solid cake is discarded. As before, the mass of the entrained liquid in the solid cake is \(15 \%\) of the mass of liquid charged to the mixer. 4\. The contents of the filtrate holding tank are filtered to remove the carried-over spent leaf, and the wet cake is pressed to recover entrained liquid, which is combined with the filtrate. A negligible amount of liquid remains in the wet cake. The filtrate, which contains \(\mathrm{D}, \mathrm{E},\) and \(\mathrm{W},\) is pumped to an extraction unit (another mixer). 5\. In the extraction unit, the alcohol-water-drug solution is contacted with another solvent (F), which is almost but not completely immiscible with ethanol and water. Essentially all of the drug (D) is extracted into the second solvent, from which it is eventually separated by a process of no concern in this problem. Some ethanol but no water is also contained in the extract. The solution from which the drug has been extracted (the raffinate) contains \(13.0 \mathrm{wt} \% \mathrm{E}, 1.5 \% \mathrm{F},\) and \(85.5 \%\) W. It is fed to a stripping column for recovery of the ethanol. 6\. The feeds to the stripping column are the solution just described and steam. The two streams are fed in a ratio such that the overhead product stream from the column contains \(20.0 \mathrm{wt} \% \mathrm{E}\) and \(2.6 \% \mathrm{F},\) and the bottom product stream contains \(1.3 \mathrm{wt} \% \mathrm{E}\) and the balance \(\mathrm{W}\). Draw and label a flowchart of the process, taking as a basis one batch of leaf processed. Then calculate (a) the masses of the components of the filtrate holding tank. (b) the masses of the components \(D\) and \(E\) in the extract stream leaving the extraction unit. (c) the mass of steam fed to the stripping column, and the masses of the column overhead and bottoms products.

\- An equimolar liquid mixture of benzene and toluene is separated into two product streams by distillation. A process flowchart and a somewhat oversimplified description of what happens in the process follow: Inside the column a liquid stream flows downward and a vapor stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving the top of the column, which contains 97 mole\% benzene, is completely condensed and split into two equal fractions: one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains \(89.2 \%\) of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which \(45 \%\) of it is vaporized. The vapor generated in the reboiler (the boilup) is recycled to become the rising vapor stream in the column, and the residual reboiler liquid is taken off as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation $$\frac{y_{\mathrm{B}} /\left(1-y_{\mathrm{B}}\right)}{x_{\mathrm{B}} /\left(1-x_{\mathrm{B}}\right)}=2.25$$ where \(y_{\mathrm{B}}\) and \(x_{\mathrm{B}}\) are the mole fractions of benzene in the vapor and liquid streams, respectively. (a) Take a basis of 100 mol fed to the column. Draw and completely label a flowchart, and for each of four systems (overall process, column, condenser, and reboiler), do the degree-of-freedom analysis and identify a system with which the process analysis might appropriately begin (one with zero degrees of freedom). (b) Write in order the equations you would solve to determine all unknown variables on the flowchart, circling the variable for which you would solve in each equation. Do not do the calculations in this part. (c) Calculate the molar amounts of the overhead and bottoms products, the mole fraction of benzene in the bottoms product, and the percentage recovery of toluene in the bottoms product \((100 \times\) moles toluene in bottoms/mole toluene in feed).
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