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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

Draw and label the given streams and derive expressions for the indicated quantities in terms of labeled variables. The solution of Part (a) is given as an illustration. (a) A continuous stream contains 40.0 mole\% benzene and the balance toluene. Write expressions for the molar and mass flow rates of benzene, \(\dot{n}_{\mathrm{B}}\left(\operatorname{mol} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right)\) and \(\dot{m}_{\mathrm{B}}\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right),\) in terms of the total molar flow rate of the stream, \(\dot{n}(\mathrm{mol} / \mathrm{s})\) (b) The feed to a batch process contains equimolar quantities of nitrogen and methane. Write an expression for the kilograms of nitrogen in terms of the total moles \(n(\) mol) of this mixture. (c) A stream containing ethane, propane, and butane has a mass flow rate of \(100.0 \mathrm{g} / \mathrm{s}\). Write an expression for the molar flow rate of ethane, \(\dot{n}_{\mathrm{E}}\left(\text { Ib-mole } \mathrm{C}_{2} \mathrm{H}_{6} / \mathrm{h}\right)\), in terms of the mass fraction of this species, \(x_{\mathrm{E}}\). (d) A continuous stream of humid air contains water vapor and dry air, the latter containing approximately 21 mole \(\% \mathrm{O}_{2}\) and \(79 \% \mathrm{N}_{2}\). Write expressions for the molar flow rate of \(\mathrm{O}_{2}\) and for the mole fractions of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2}\) in the gas in terms of \(\dot{n}_{1}\left(\mathrm{lb}-\mathrm{mole} \mathrm{H}_{2} \mathrm{O} / \mathrm{s}\right)\) and \(\dot{n}_{2}(\text { lb- mole dry air/s })\) (e) The product from a batch reactor contains \(\mathrm{NO}, \mathrm{NO}_{2},\) and \(\mathrm{N}_{2} \mathrm{O}_{4} .\) The mole fraction of \(\mathrm{NO}\) is 0.400. Write an expression for the gram-moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) in terms of \(n(\mathrm{mol}\) mixture) and \(y_{\mathrm{NO}_{2}}\left(\operatorname{mol} \mathrm{NO}_{2} / \mathrm{mol}\right)\)

Ammonia is oxidized to nitric oxide in the following reaction: $$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$$ (a) Calculate the ratio (lb-mole \(\mathrm{O}_{2}\) react/lb-mole NO formed). (b) If ammonia is fed to a continuous reactor at a rate of \(100.0 \mathrm{kmol} \mathrm{NH}_{3} / \mathrm{h}\), what oxygen feed rate (kmol/h) would correspond to 40.0\% excess O_? (c) If \(50.0 \mathrm{kg}\) of ammonia and \(100.0 \mathrm{kg}\) of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

A fuel oil is fed to a furnace and burned with \(25 \%\) excess air. The oil contains \(87.0 \mathrm{wt} \% \mathrm{C}, 10.0 \% \mathrm{H},\) and 3.0\% S. Analysis of the furnace exhaust gas shows only \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}, \mathrm{SO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\). The sulfur dioxide emission rate is to be controlled by passing the exhaust gas through a scrubber, in which most of the \(\mathrm{SO}_{2}\) is absorbed in an alkaline solution. The gases leaving the scrubber (all of the \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{CO}_{2}\), and some of the \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{SO}_{2}\) entering the unit) pass out to a stack. The scrubber has a limited capacity, however, so that a fraction of the furnace exhaust gas must be bypassed directly to the stack. At one point during the operation of the process, the scrubber removes \(90 \%\) of the \(\mathrm{SO}_{2}\) in the gas fed to it, and the combined stack gas contains 612.5 ppm (parts per million) \(\mathrm{SO}_{2}\) on a dry basis; that is, every million moles of dry stack gas contains 612.5 moles of \(\mathrm{SO}_{2}\). Calculate the fraction of the exhaust bypassing the scrubber at this moment.

If the percentage of fuel in a fuel-air mixture falls below a certain value called the lower flammability limit (LFL), which sometimes is referred to as the lower explosion limit (LEL), the mixture cannot be ignited. In addition there is an upper flammability limit (UFL), which also is known as the upper explosion limit (UEL). For example, the LFL of propane in air is 2.3 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) and the UFL is \(9.5 \%^{14}\). If the percentage of propane in a propane-air mixture is greater than \(2.3 \%\) and less than \(9.5 \%,\) the gas mixture can ignite if it is exposed to a flame or spark. A mixture of propane in air containing 4.03 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) (fuel gas) is the feed to a combustion furnace. If there is a problem in the furnace, a stream of pure air (dilution air) is added to the fuel mixture prior to the furnace inlet to make sure that ignition is not possible. (a) Draw and label a flowchart of the fuel gas-dilution air mixing unit, presuming that the gas entering the furnace contains propane at the LFL, and do the degree-of-freedom analysis. (b) If propane flows at a rate of \(150 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8} / \mathrm{s}\) in the original fuel-air mixture, what is the minimum molar flow rate of the dilution air? (c) How would the actual dilution air feed rate probably compare with the value calculated in Part (b)? (>, \(<,=\) ) Explain.

Methane reacts with chlorine to produce methyl chloride and hydrogen chloride. Once formed, the methyl chloride may undergo further chlorination to form methylene chloride ( \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) ), chloroform, and carbon tetrachloride. A methyl chloride production process consists of a reactor, a condenser, a distillation column, and an absorption column. A gas stream containing 80.0 mole \(\%\) methane and the balance chlorine is fed to the reactor. In the reactor a single-pass chlorine conversion of essentially \(100 \%\) is attained, the mole ratio of methyl chloride to methylene chloride in the product is \(5: 1,\) and negligible amounts of chloroform and carbon tetrachloride are formed. The product stream flows to the condenser. Two streams emerge from the condenser: the liquid condensate, which contains essentially all of the methyl chloride and methylene chloride in the reactor effluent, and a gas containing the methane and hydrogen chloride. The condensate goes to the distillation column in which the two component species are separated. The gas leaving the condenser flows to the absorption column where it contacts an aqueous solution. The solution absorbs essentially all of the HCl and none of the \(\mathrm{CH}_{4}\) in the feed. The liquid leaving the absorber is pumped elsewhere in the plant for further processing, and the methane is recycled to join the fresh feed to the process (a mixture of methane and chlorine). The combined stream is the feed to the reactor. (a) Choose a quantity of the reactor feed as a basis of calculation, draw and label a flowchart, and determine the degrees of freedom for the overall process and each single unit and stream mixing point. Then write in order the equations you would use to calculate the molar flow rate and molar composition of the fresh feed, the rate at which HCI must be removed in the absorber, the methyl chloride production rate, and the molar flow rate of the recycle stream. Do no calculations. (b) Calculate the quantities specified in Part (a), either manually or with an equation-solving program. (c) What molar flow rates and compositions of the fresh feed and the recycle stream are required to achieve a methyl chloride production rate of \(1000 \mathrm{kg} / \mathrm{h} ?\)
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