/*! 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 56 The reaction between ethylene an... [FREE SOLUTION] | 91Ó°ÊÓ

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The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product stream is analyzed and found to contain 51.7 mole \(\% \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\) and 17.3\% HBr. The feed to the reactor contains only ethylene and hydrogen bromide. Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is \(165 \mathrm{mol} / \mathrm{s}\), what is the extent of reaction?

Short Answer

Expert verified
The fractional conversion of the limiting reactant, HBr, is approximately 64.2%. The reactant C2H4 is in excess by about 35.8%. The extent of the reaction is 51.15 mol/s.

Step by step solution

01

Determining the limiting reactant

Since the reaction between ethylene (\(C_{2}H_{4}\)) and hydrogen bromide (\(HBr\)) leads to the formation of ethyl bromide(\(C_{2}H_{5}Br\)), the stoichiometric reaction is as follows: \(C_{2}H_{4} + HBr \rightarrow C_{2}H_{5}Br\). We can determine the limiting reactant by comparing the mole fractions from the analysis of the product stream and the initial feed. Since the product stream contains \(17.3\% HBr\), it would mean that \(HBr\) is the limiting reactant because it has not been completely consumed.
02

Calculating fractional conversion of limiting reactant

The fractional conversion of the limiting reactant can be calculated using the equation: \[conversion = \frac{initial\ moles - final\ moles}{initial\ moles}\] Here, as \(HBr\) is the limiting reactant, the initial moles of \(HBr\) would have been equal to the total moles minus moles of \(C_{2}H_{4}\). Therefore, the conversion = \(1 - \frac{0.173}{0.483}\) = \(0.642\). Thus, approximately \(64.2\%\) of the limiting reactant \(HBr\) is converted to the product.
03

Determining the excess of the other reactant

The percentage by which the other reactant is in excess is given by the equation: \[ excess = \frac{initial\ moles - converted\ moles}{initial\ moles} * 100\% \] Here, \(C_{2}H_{4}\) is the other reactant. Using the stoichiometry of the reaction, we should observe that the moles of \(C_{2}H_{4}\) should be same as the consumed moles of \(HBr\). Hence, the excess of \(C_{2}H_{4}\) = \(1 - conversion\) = \(1 - 0.642\) = \(35.8\%\) Therefore, approximately \(35.8\%\) of \(C_{2}H_{4}\) is in excess.
04

Determining the extent of reaction

The extent of the reaction can be determined by using the equation: \[extent\ of\ reaction = \frac{\Delta n}{\nu}\] where \(\Delta n\) is the change in mole numbers and \(\nu\) is the stoichiometric coefficient, which is equal to 1 for both reactants in our case. Here, the mole numbers change by the number of moles of the limiting reactant initially present - number of moles of limiting reactant finally present = \(0.483*165 - 0.173*165\) = \(51.15\) mol/s. Hence, the extent of reaction = \(\frac{51.15}{1}\) = \(51.15\) mol/s.

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Key Concepts

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

Limiting Reactant
Understanding the concept of a limiting reactant is crucial in chemical reactions, as it dictates the maximum amount of product that can be formed. This reactant is the one that will be completely used up first, and thus limits the progression of the reaction. In a chemical equation, the mole ratio of reactants is defined by the balanced equation, and the reactant present in the lesser amount, according to this ratio, is the limiting one.

Consider a simple reaction where ingredient A reacts with ingredient B to form product C. If you run out of A before B, then A is the limiting reactant. Similarly, in our given exercise, hydrogen bromide (HBr) was identified as the limiting reactant because the product stream contained unreacted HBr, suggesting all the ethylene (C2H4) could not react as there wasn't enough HBr available.
Fractional Conversion Calculation
The fractional conversion calculation quantifies the proportion of a reactant that has been used up in the reaction; it is a measure of the reaction's progress. The calculation is straightforward: it's the initial amount of the reactant minus the amount that remains, divided by the initial amount. Expressed as a fraction (hence, 'fractional'), it can also be converted into a percentage.

In our example, to calculate the fractional conversion of hydrogen bromide, you subtract the final moles of HBr from the initial moles (which can be found using the mole fraction of HBr in the feed) and then divide by the initial moles. Mathematically, this is represented by the formula \[\text{Conversion} = \frac{\text{initial moles} - \text{final moles}}{\text{initial moles}}\] When successfully calculated, this gives you direct insight into how much of your limiting reactant was consumed during the chemical reaction.
Excess Reactant Percentage
While the limiting reactant is completely used up, the excess reactant is the one left over after the reaction goes to completion. Knowing the percentage of the excess reactant is helpful for understanding the reaction efficiency and planning for resource management in industrial processes.

The excess reactant percentage is found by the formula \[ \text{Excess} = \frac{\text{initial moles} - \text{converted moles}}{\text{initial moles}} \times 100\% \] In this formula, 'converted moles' is the number of moles of the excess reactant that actually participated in the reaction, which can be deduced from the stoichiometry of the chemical equation. In our reaction, the ethylene was in excess; thus, we could determine how much ethylene did not react and remained in the reactor.
Extent of Reaction
Calculating the extent of reaction gives us another piece of important information, it represents the quantity in moles of a product formed or reactant consumed in a given reaction. The extent is calculated by determining the change in moles (\(\text{Δn}\)) of reactants or products, and dividing by the stoichiometric coefficient (\(u\)), which is the 'balanced equation coefficient' for each substance in the reaction.

For the reaction between ethylene and hydrogen bromide, the stoichiometric coefficients for both reactants are equal to 1 (as per the balanced equation), simplifying the calculation. The extent of reaction tells us how far the reaction has gone in terms of molar consumption and is directly related to the flow rate of the product stream in continuous reactors, like in the case of our exercise.

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Most popular questions from this chapter

Carbon nanotubes (CNT) are among the most versatile building blocks in nanotechnology. These unique pure carbon materials resemble rolled-up sheets of graphite with diameters of several nanometers and lengths up to several micrometers. They are stronger than steel, have higher thermal conductivities than most known materials, and have electrical conductivities like that of copper but with higher currentcarrying capacity. Molecular transistors and biosensors are among their many applications. While most carbon nanotube research has been based on laboratory-scale synthesis, commercial applications involve large industrial-scale processes. In one such process, carbon monoxide saturated with an organo-metallic compound (iron penta-carbonyl) is decomposed at high temperature and pressure to form CNT, amorphous carbon, and CO_. Each "molecule" of CNT contains roughly 3000 carbon atoms. The reactions by which such molecules are formed are: In the process to be analyzed, a fresh feed of CO saturated with \(\mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{v})\) contains \(19.2 \mathrm{wt} \%\) of the latter component. The feed is joined by a recycle stream of pure CO and fed to the reactor, where all of the iron penta-carbonyl decomposes. Based on laboratory data, \(20.0 \%\) of the CO fed to the reactor is converted, and the selectivity of CNT to amorphous carbon production is (9.00 kmol CNT/kmol C). The reactor effluent passes through a complex separation process that yields three product streams: one consists of solid \(\mathrm{CNT}, \mathrm{C},\) and \(\mathrm{Fe} ;\) a second is \(\mathrm{CO}_{2} ;\) and the third is the recycled \(\mathrm{CO}\). You wish to determine the flow rate of the fresh feed (SCM/h), the total CO_ generated in the process ( \(\mathrm{kg} / \mathrm{h}\) ), and the ratio (kmol CO recycled/kmol CO in fresh feed). (a) Take a basis of \(100 \mathrm{kmol}\) fresh feed. Draw and fully label a process flow chart and do degree-offreedom analyses for the overall process, the fresh-feed/recycle mixing point, the reactor, and the separation process. Base the analyses for reactive systems on atomic balances. (b) Write and solve overall balances, and then scale the process to calculate the flow rate (SCM/h) of fresh feed required to produce \(1000 \mathrm{kg} \mathrm{CNT} / \mathrm{h}\) and the mass flow rate of \(\mathrm{CO}_{2}\) that would be produced. (c) In your degree-of-freedom analysis of the reactor, you might have counted separate balances for C (atomic carbon) and O (atomic oxygen). In fact, those two balances are not independent, so one but not both of them should be counted. Revise your analysis if necessary, and then calculate the ratio (kmol CO recycled/kmol CO in fresh feed). (d) Prove that the atomic carbon and oxygen balances on the reactor are not independent equations.

A liquid-phase chemical reaction \(\mathrm{A} \rightarrow \mathrm{B}\) takes place in a well-stirred tank. The concentration of \(\mathrm{A}\) in the feed is \(C_{\mathrm{A} 0}\left(\operatorname{mol} / \mathrm{m}^{3}\right),\) and that in the tank and outlet stream is \(C_{\mathrm{A}}\left(\mathrm{mol} / \mathrm{m}^{3}\right) .\) Neither concentration varies with time. The volume of the tank contents is \(V\left(\mathrm{m}^{3}\right)\) and the volumetric flow rate of the inlet and outlet streams is \(\dot{V}\left(\mathrm{m}^{3} / \mathrm{s}\right)\). The reaction rate (the rate at which \(\mathrm{A}\) is consumed by reaction in the tank) is given by the expression $$r(\text { mol } A \text { consumed } / \mathrm{s})=k V C_{\mathrm{A}}$$ (a) Is this process continuous, batch, or semibatch? Is it transient or steady-state? (b) What would you expect the reactant concentration \(C_{\mathrm{A}}\) to equal if \(k=0\) (no reaction)? What should it approach if \(k \rightarrow \infty\) (infinitely rapid reaction)? (c) Write a differential balance on \(A,\) stating which terms in the general balance equation (accumulation = input + generation - output - consumption) you discarded and why you discarded them. Use the balance to derive the following relation between the inlet and outlet reactant concentrations: $$C_{\mathrm{A}}=\frac{C_{\mathrm{A} 0}}{1+k V / \dot{V}}$$ Verify that this relation predicts the results in Part (b).

Methanol is formed from carbon monoxide and hydrogen in the gas-phase reaction The mole fractions of the reactive species at equilibrium satisfy the relation where \(P\) is the total pressure (atm), \(K_{c}\) the reaction equilibrium constant (atm \(^{-2}\) ), and \(T\) the temperature (K). The equilibrium constant \(K_{c}\) equals 10.5 at 373 K, and \(2.316 \times 10^{-4}\) at \(573 \mathrm{K}\). A semilog plot of \(K_{\mathrm{c}}\) (logarithmic scale) versus 1/ \(T\) (rectangular scale) is approximately linear between \(T=300 \mathrm{K}\) and \(T=600 \mathrm{K}\) (a) Derive a formula for \(K_{\mathrm{c}}(T),\) and use it to show that \(K_{\mathrm{e}}(450 \mathrm{K})=0.0548 \mathrm{atm}^{-2}\) (b) Write expressions for \(n_{A}, n_{B},\) and \(n_{C}\) (gram-moles of each species), and then \(y_{A}, y_{B},\) and \(y_{C},\) in terms of \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0},\) and \(\xi,\) the extent of reaction. Then derive an equation involving only \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0}, P, T,\) and \(\xi_{e},\) where \(\xi_{e}\) is the extent of reaction at equilibrium. (c) Suppose you begin with equimolar quantities of CO and \(\mathrm{H}_{2}\) and no \(\mathrm{CH}_{3} \mathrm{OH}\), and the reaction proceeds to equilibrium at 423 K and 2.00 atm. Calculate the molar composition of the product ( \(y_{\mathrm{A}}\), \(\left.y_{\mathrm{B}}, \text { and } y_{\mathrm{C}}\right)\) and the fractional conversion of \(\mathrm{CO}\) (d) The conversion of CO and \(\mathrm{H}_{2}\) can be enhanced by removing methanol from the reactor while leaving unreacted CO and \(\mathrm{H}_{2}\) in the vessel. Review the equations you derived in solving Part (c) and determine any physical constraints on \(\xi_{c}\) associated with \(n_{\mathrm{A} 0}=n_{\mathrm{B} 0}=1\) mol. Now suppose that 90\% of the methanol is removed from the reactor as it is produced; in other words, only 10\% of the methanol formed remains in the reactor. Estimate the fractional conversion of CO and the total gram moles of methanol produced in the modified operation. (e) Repeat Part (d), but now assume that \(n_{\mathrm{B} 0}=2\) mol. Explain the significant increase in fractional conversion of CO. (f) Write a set of equations for \(y_{\mathrm{A}}, y_{\mathrm{B}}, y_{\mathrm{C}},\) and \(f_{\mathrm{A}}\) (the fractional conversion of \(\mathrm{CO}\) ) in terms of \(y_{\mathrm{A} 0}, y_{\mathrm{B} 0}, T,\) and \(P(\) the reactor temperature and pressure at equilibrium). Enter the equations in an equation-solving program. Check the program by running it for the conditions of Part (c), then use it to determine the effects on \(f_{\mathrm{A}}\) (increase, decrease, or no effect) of separately increasing, (i) the fraction of \(\mathrm{CH}_{3} \mathrm{OH}\) in the feed, (ii) temperature, and (iii) pressure.

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?)

The hormone estrogen is produced in the ovaries of females and elsewhere in the body in men and postmenopausal women, and it is also administered in estrogen replacement therapy, a common treatment for women who have undergone a hysterectomy. Unfortunately, it also binds to estrogen receptors in breast tissue and can activate cells to become cancerous. Tamoxifen is a drug that also binds to estrogen receptors but does not activate cells, in effect blocking the receptors from access to estrogen and inhibiting the growth of breast-cancer cells. Tamoxifen is administered in tablet form. In the manufacturing process, a finely ground powder contains tamoxifen (tam) and two inactive fillers- -lactose monohydrate (lac) and corn starch (cs). The powder is mixed with a second stream containing water and suspended solid particles of polyvinylpymolidone (pvp) binder, which keeps the tablets from easily crumbling. The slurry leaving the mixer goes to a dryer, in which 94.2\% of the water fed to the process is vaporized. The wet powder leaving the dryer contains 8.80 wr\% tam, 66.8\% lac, 21.4\% cs, 2.00\% pvp, and 1.00\% water. After some additional processing, the powder is molded into tablets. To produce a hundred thousand tablets, 17.13 kg of wet powder is required. (a) Taking a basis of 100,000 tablets produced, draw and label a process flowchart, labeling masses of individual components rather than total masses and component mass fractions. It is unnecessary to label the stream between the mixer and the dryer. Carry out a degree-of-freedom analysis of the overall two-unit process. (b) Calculate the masses and compositions of the streams that must enter the mixer to make 100,000 tablets. (c) Why was it unnecessary to label the stream between the mixer and the dryer? Under what circumstances would it have been necessary? (d) Go back to the flowchart of Part (a). Without using the mass of the wet powder (17.13 kg) or any of the results from Part (b) in your calculations, determine the mass fractions of the stream components in the powder fed to the mixer and verify that they match your solution to Part (b). (Hint: Take a basis of \(100 \mathrm{kg}\) of wet powder.) (e) Suppose a student does Part (d) before Part (b), and re-labels the powder feed to the mixer on the flowchart of Part (a) with an unknown total mass ( \(m_{1}\) ) and the three now known mole fractions. (Sketch the resulting flowchart.) The student then does a degree-of-freedom analysis, counts four unknowns (the masses of the powder, pvp, and water fed to the mixer, and the mass of water evaporated in the dryer), and six equations (five material balances for five species and the percentage evaporation), for a net of -2 degrees of freedom. since there are more equations than unknowns, it should not be possible to get a unique solution for the four unknowns. Nevertheless, the student writes four equations, solves for the four unknowns, and verifies that all of the balance equations are satisfied. There must have been a mistake in the degree-of-freedom calculation. What was it?

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