/*! 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 62 Oxygen consumed by a living orga... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Oxygen consumed by a living organism in aerobic reactions is used in adding mass to the organism and/or the production of chemicals and carbon dioxide. since we may not know the molecular compositions of all species in such a reaction, it is common to define the ratio of moles of \(\mathrm{CO}_{2}\) produced per mole of \(\mathrm{O}_{2}\) consumed as the respiratory quotient, \(R Q,\) where $$R Q=\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{O}_{2}}}\left(\text { or } \frac{\dot{n}_{\mathrm{CO}_{2}}}{\dot{n}_{\mathrm{O}_{2}}}\right)$$ since it generally is impossible to predict values of \(R Q\), they must be determined from operating data. Mammalian cells are used in a bioreactor to convert glucose to glutamic acid by the reaction $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+a \mathrm{NH}_{3}+b \mathrm{O}_{2} \rightarrow p \mathrm{C}_{5} \mathrm{H}_{9} \mathrm{NO}_{4}+q \mathrm{CO}_{2}+r \mathrm{H}_{2} \mathrm{O}$$ The feed to the bioreactor comprises \(1.00 \times 10^{2} \mathrm{mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} / \mathrm{day}, 1.20 \times 10^{2} \mathrm{mol} \mathrm{NH}_{3} / \mathrm{day},\) and \(1.10 \times\) \(10^{2}\) mol \(\mathrm{O}_{2} /\) day. Data on the system show that \(R Q=0.45 \mathrm{mol} \mathrm{CO}_{2}\) produced/mol \(\mathrm{O}_{2}\) consumed. (a) Determine the five stoichiometric coefficients and the limiting reactant. (b) Assuming that the limiting reactant is consumed completely, calculate the molar and mass flow rates of all species leaving the reactor and the fractional conversions of the non-limiting reactants.

Short Answer

Expert verified
The five stoichiometric coefficients are a: 38.08, b: 2.22, p: 1.3, q: 49.5, and r: 135.37. The limiting reactant is Oxygen (O2). Molar flow rates for each species are as follows: C6H12O6: 0 mol/day, NH3: 41.92 mol/day, O2: 0 mol/day, C5H9NO4: 143 mol/day, CO2: 49.5 mol/day, H2O: 135.37 mol/day. The fractional conversions for each non-limiting reactant are as follows: NH3: 0.651 mol/day, Glucose: 1 mol/day.

Step by step solution

01

Finding stoichiometric coefficients

Firstly calculate stoichiometric coefficients 'p' and 'q'. Start out by using the given Respiratory Quotient (RQ). Rearrange the RQ formula to solve for \(q\): \(RQ = n_{CO2} / n_{O2}\) can be written as \(q = RQ * n_{O2}\). Substituting known values gives: \(q = 0.45 * 110 = 49.5 mol/day\). Second, the reaction formula also equates carbon atoms on both sides. This gives \(6 = p*5 + q\). Substituting the known value for 'q' in this equation yields 'p' as \(56 = p*5 + 49.5 -> p = (56 - 49.5) / 5 = 1.3\).
02

Identifying the limiting reactant

The limiting reactant is the one which is consumed the most in relation to its stoichiometric coefficient. Here, the coefficient relates to 'a' and 'b'. Using the equation for reaction, we get \(C6H12O6 + aNH3 + bO2 -> 1.3C5H9NO4 + 49.5CO2 + rH2O\). By comparing the ratios of reactants to their stoichiometric coefficients, it can be seen that the O2 is limiting since it has the lowest amount of 110/0.45 = 244.44. So, \(a = 49.5 / 1.3 = 38.08\) and \(b = 1 / RQ = 1 / 0.45 = 2.22\). 'r' can be calculated by balancing Hydrogen and Oxygen on both sides of the reaction.
03

Molar and Mass Flow Rates, Fractional Conversions

Assuming that the limiting reactant (O2) is completely consumed, the mass flow rates and fractional conversion of the non-limiting reactants (NH3 and Glucose) can be calculated. Firstly, the flow rates leaving the reactor can be calculated by the reaction stoichiometry. For CO2, it would be \(0.45 * 110\). For C5H9NO4, it would be \(1.3 * 110\) and for NH3, \(38.08 * 110\). For glucose, \(110 - (1*110) = 0\) since it is completely consumed. Lastly, the fractional conversion of NH3 can be calculated as \(1 - (-consuming flows / original feeding flows)\).
04

Calculating 'r'

Balance the number of Hydrogen and atoms to get \(r\). On the left-hand side of the equation, we have \(12 + 3a\) Hydrogen atoms and \(6 + 3b\) Oxygen atoms. On the right-hand side of the equation, we have \(9*1.3 + 2q + 2r\) Hydrogen atoms and \(4*1.3 + 2q + r\) Oxygen atoms. Equating both sides for Hydrogen and Oxygen balances, we obtain ordinary equations which we can solve to get \(r = 135.37\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Aerobic Reactions
In the realm of biological and chemical processes, aerobic reactions are essential for sustaining life. These reactions occur in the presence of oxygen and are crucial for the conversion of substrates like glucose into energy, carbon dioxide, and water.
When organisms, particularly aerobic organisms, metabolize glucose, they undergo a series of reactions where oxygen is used to break down glucose, often producing carbon dioxide and water as byproducts. This process of aerobic respiration is pivotal as it provides energy that cells need to function effectively.
In the exercise, the focus is on the transformation of glucose into compounds like glutamic acid using oxygen as a reactant.
This understanding is key because it underlines why monitoring and optimizing oxygen consumption is critical in bioreactor operations, ensuring the desired product is maximized while maintaining metabolic balance.
Stoichiometry
Stoichiometry is a concept that depicts the quantitative nature of chemical reactions, defined by the balancing of molar amounts of reactants and products.
In this exercise, we use stoichiometry to establish the relationships between the quantities of glucose, ammonia, and oxygen and the resulting products such as carbon dioxide and glutamic acid.
By calculating stoichiometric coefficients, students can understand how much of each reactant is needed or produced to maintain balance within a chemical equation. For instance, knowing the stoichiometric coefficient of a reactant helps determine how it proportionally affects the products formed.
The stoichiometric coefficients allow us to interpret how efficiently the reaction occurs and if the reactants are fully utilized, which is directly applied to practical settings like managing reactants within a bioreactor.
Limiting Reactant
The concept of a limiting reactant is critical in chemical reactions as it dictates the extent to which a reaction can proceed. It is the substance that gets completely consumed first, thus limiting the formation of products.
In our exercise, identifying oxygen as the limiting reactant was paramount. Why? Because it allows us to calculate the maximum amount of products that can be formed.
Without determining the limiting reactant, any calculation regarding yields and efficiency would be skewed, potentially leading to a misunderstanding of the system's behavior.
This concept not only helps in academic exercises but also in industrial processes, where ensuring that limiting reactants are adequately supplied can influence the cost and efficiency of chemical production.
Bioreactor
Bioreactors are vessels or devices used to carry out a chemical process involving organisms or biochemically active substances derived from such organisms. They are integral in various biotechnological applications, especially in the production of pharmaceuticals, biofuels, and food products.
In the given exercise, the bioreactor functions as the setting where mammalian cells convert glucose into glutamic acid under controlled aerobic conditions.
Understanding how to control the parameters within a bioreactor, such as oxygen levels, is crucial. This is particularly significant when managing metabolic rates and ensuring the efficient production of desired substances with minimal waste.
Therefore, bioreactors exemplify the practical application of aerobic reactions and chemical balancing through stoichiometry to maximize output and improve process efficiency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A gas contains 75.0 wt\% methane, \(10.0 \%\) ethane, \(5.0 \%\) ethylene, and the balance water. (a) Calculate the molar composition of this gas on both a wet and a dry basis and the ratio (mol \(\mathrm{H}_{2} \mathrm{O} /\) mol dry gas). (b) If \(100 \mathrm{kg} / \mathrm{h}\) of this fuel is to be burned with \(30 \%\) excess air, what is the required air feed rate (kmol/ h)? How would the answer change if the combustion were only \(75 \%\) complete?

Inside a distillation column (see Problem 4.8), a downward-flowing liquid and an upward-flowing vapor maintain contact with each other. For reasons we will discuss in greater detail in Chapter \(6,\) the vapor stream becomes increasingly rich in the more volatile components of the mixture as it moves up the column, and the liquid stream is enriched in the less volatile components as it moves down. The vapor leaving the top of the column goes to a condenser. A portion of the condensate is taken off as a product (the overhead product), and the remainder (the reflux) is returned to the top of the column to begin its downward journey as the liquid stream. The condensation process can be represented as shown below: A distillation column is being used to separate a liquid mixture of ethanol (more volatile) and water (less volatile). A vapor mixture containing 89.0 mole \(\%\) ethanol and the balance water enters the overhead condenser at a rate of \(100 \mathrm{lb}\) -mole/h. The liquid condensate has a density of \(49.01 \mathrm{b}_{\mathrm{m}} / \mathrm{ft}^{3},\) and the reflux ratio is \(3 \mathrm{lb}_{\mathrm{m}}\) reflux/lb \(_{\mathrm{m}}\) overhead product. When the system is operating at steady state, the tank collecting the condensate is half full of liquid and the mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) is 10.0 minutes. Determine the overhead product volumetric flow rate (ft \(^{3}\) /min) and the condenser tank volume (gal).

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?

Acetylene is hydrogenated to form ethane. The feed to the reactor contains \(1.50 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}\) (a) Calculate the stoichiometric reactant ratio (mol \(\mathrm{H}_{2}\) react/mol \(\mathrm{C}_{2} \mathrm{H}_{2}\) react) and the yield ratio (kmol \(\mathbf{C}_{2} \mathbf{H}_{6}\) formed/kmol \(\mathbf{H}_{2}\) react (b) Determine the limiting reactant and calculate the percentage by which the other reactant is in excess. (c) Calculate the mass feed rate of hydrogen ( \(\mathrm{kg} / \mathrm{s}\) ) required to produce \(4 \times 10^{6}\) metric tons of ethane per year, assuming that the reaction goes to completion and that the process operates for 24 hours a day, 300 days a year. (d) There is a definite drawback to running with one reactant in excess rather than feeding the reactants in stoichiometric proportion. What is it? [Hint: In the process of Part (c), what does the reactor effluent consist of and what will probably have to be done before the product ethane can be sold or used?]

Seawater containing 3.50 wt\% salt passes through a series of 10 evaporators. Roughly equal quantities of water are vaporized in each of the 10 units and then condensed and combined to obtain a product stream of fresh water. The brine leaving each evaporator but the tenth is fed to the next evaporator. The brine leaving the tenth evaporator contains \(5.00 \mathrm{wt} \%\) salt. (a) Draw a flowchart of the process showing the first, fourth, and tenth evaporators. Label all the streams entering and leaving these three evaporators. (b) Write in order the set of equations you would solve to determine the fractional yield of fresh water from the process \(\left(\mathrm{kg} \mathrm{H}_{2} \mathrm{O} \text { recovered } / \mathrm{kg} \mathrm{H}_{2} \mathrm{O}\) in process feed) and the weight percent of salt in the \right. solution leaving the fourth evaporator. Each equation you write should contain no more than one previously undetermined variable. In each equation, circle the variable for which you would solve. Do not do the calculations. (c) Solve the equations derived in Part (b) for the two specified quantities. (d) The problem statement made no mention of the disposition of the 5 wt\% effluent from the tenth evaporator. Suggest two possibilities for its disposition and describe any environmental concerns that might need to be considered.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.