Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Ozone is a reactive gas that has been associated with respiratory illness and decreased lung function. The following reactions are involved in ozone formation [D. Allen and D. Shonnard, Green Engineering (Upper Saddle River, N.J.: Prentice Hall, 2002)]. \(\mathrm{NO}_{2}\) is primarily generated by combustion in the automobile engine. (a) Show that the steady-state concentration of ozone is directly proportional to \(\mathrm{NO}_{2}\) and inversely proportional to NO. (b) Drive an equation for the concentration of ozone in solely in terms of the initial concentrations \(C_{\mathrm{NO}, 0}, C_{\mathrm{VO}, 0},\) and \(\mathrm{O}_{3,0}\) and the rate law parameters. (c) In the absence of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), the rate law for ozone generation is $$-r_{0,3}=\frac{k\left(\mathrm{O}_{2}\right)(\mathrm{M})}{\left(\mathrm{O}_{2}\right)(\mathrm{M})+k^{\prime}\left(\mathrm{O}_{3}\right)}$$ Suggest a mechanism. (d) List ways you can work this problem incorrectly. (e) How could you make this problem more difficult?

Short Answer

Expert verified
a) Ozone concentration is directly proportional to NO₂ concentration and inversely proportional to NO concentration: \(O_3 = \frac{k_2 [NO_2]}{k_1 [NO]}\). b) The equation for ozone concentration in terms of initial concentrations and rate law parameters is: \[C_{O_3,ss}=\frac{k_2}{k_1}\left[C_{NO_2,0}+(C_{NO,0} - \xi_1) - \left(C_{O_{3,0}} + \xi_1 - \xi_2\right)\right]- \frac{C_{O_{3,0}}}{C_{NO,0}}\] c) A possible mechanism in the absence of NO and NO₂ involves the dissociation of molecular oxygen (O₂) and formation and dissociation of ozone (O₃) with the presence of a third body (M). d) Some ways to work this problem incorrectly include not considering steady-state conditions, not accounting for initial concentrations or rate law parameters, and not applying the correct mechanism in part (c). e) To make this problem more difficult, introduce reaction intermediates, temperature dependence, catalysts effects, non-steady-state behaviors, or other pollutants interacting with NO and NO₂.

Step by step solution

01

a) Proportionality of Ozone Concentration to NO₂ and NO

Begin by writing the reactions involved in ozone formation: Reaction 1: \(NO + O_3 \rightarrow NO_2 + O_2\) (Ozone reacts with nitric oxide) Reaction 2: \(NO_2 + O_3 \rightarrow NO + 2O_2\) (Nitrogen dioxide reacts with ozone) At steady state: \(r_1 = k_1 [NO][O_3] = r_2 = k_2 [NO_2][O_3]\) To find the proportionality we need to get Ozone concentration in terms of NO and NO₂ concentrations: \(O_3 = \frac{k_2 [NO_2]}{k_1 [NO]}\) The concentration of ozone is directly proportional to the concentration of NO₂ and inversely proportional to the concentration of NO.
02

b) Ozone Concentration in Terms of Initial Concentrations and Rate Law Parameters

Initially, we can write the rate laws for each reaction: Rate law for Reaction 1: \(r_1 = k_1 [NO][O_3]\) Rate law for Reaction 2: \(r_2 = k_2 [NO_2][O_3]\) At the steady state, the rate of each reaction is equal, so \(k_1 [NO][O_3] = k_2 [NO_2][O_3]\) Cancel [O₃] from both sides: \([NO]k_1=[NO_2]k_2\) Now find Ozone concentration when the reaction reaches steady-state using materials balance \(C_{O_3} = C_{O_{3,0}} + \xi_1 - \xi_2\) Where \(\xi_1\) and \(\xi_2\) are the extents of reactions 1 and 2 Divide both sides by [NO] \(C_{O_3}=\frac{k_2}{k_1}[NO_2]- \frac{C_{O_{3,0}}}{C_{NO,0}}\) At steady-state \([NO]_{ss} = C_{NO,0} - \xi_1\), \([NO_2]_{ss}=C_{NO_2,0}+\xi_1-\xi_2\) So \[C_{O_3,ss}=\frac{k_2}{k_1}\left[C_{NO_2,0}+(C_{NO,0} - \xi_1) - \left(C_{O_{3,0}} + \xi_1 - \xi_2\right)\right]- \frac{C_{O_{3,0}}}{C_{NO,0}}\] This equation represents the concentration of ozone in terms of initial concentrations and rate law parameters.
03

c) Mechanism in the Absence of NO and NO₂

The given rate law for ozone generation in the absence of NO and NO₂ is: \(-r_{0,3}=\frac{k\left(\mathrm{O}_{2}\right)(\mathrm{M})}{\left(\mathrm{O}_{2}\right)(\mathrm{M})+k^{\prime}\left(\mathrm{O}_{3}\right)}\) A possible mechanism can involve the following steps: Step 1: \(O_2 + M \rightarrow 2O + M\) (Dissociation of molecular oxygen) Step 2: \(O + O_2 + M \rightarrow O_3 + M\) (Formation of ozone) Step 3: \(O_3 + M \rightarrow O_2 + O + M\) (Dissociation of ozone) These steps can account for the given rate law and exhibit an autocatalytic behavior.
04

d) Ways to Work This Problem Incorrectly

1. Not considering the steady-state condition or the initial concentrations in deriving the ozone concentration. 2. Assuming that the rate of each reaction scales linearly with concentration. 3. Ignoring the rate law parameters in the analysis. 4. Not considering the absence of NO and NO₂ in part (c) and trying to apply the same mechanism used in the previous parts.
05

e) Making the Problem More Difficult

1. Introducing reaction intermediates or another set of reactions to the mechanism, requiring the analysis of the mechanism to be more complex. 2. Adding temperature dependence to the rate law parameters. 3. Exploring the effects of different types of catalysts on the rate of reaction and ozone concentration. 4. Considering non-steady-state behaviors and time-dependent concentration changes for the components involved. 5. Investigating the role of other pollutants in ozone formation and how they interact with NO and NO₂.

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 Vaia!

Key Concepts

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

Ozone Formation
Ozone formation is a critical topic in chemical reaction engineering, especially when considering its impact on the environment and human health. Ozone (O₃) is a reactive gas that is primarily known for its protective role in the stratosphere. However, at the ground level, ozone becomes a pollutant. Understanding how ozone forms involves examining the chemical reactions that contribute to its creation.

The initial reaction involves nitric oxide (NO) reacting with ozone to form nitrogen dioxide (NO₂) and oxygen (O₂):
  • NO + O₃ → NO₂ + O₂
Subsequently, NO₂ can react with ozone in the presence of sunlight, creating more oxygen. This cycling of NO, NO₂, and O₃ is part of what is known as the photochemical smog.

Studying ozone formation not only helps in understanding air pollution but also guides engineers in designing processes that minimize the environmental impacts of industrial emissions.
Steady-State Concentration
The concept of steady-state concentration is crucial in reaction engineering. In simple terms, a steady state occurs when the rate of input and output in a system balances out, making the concentration of reactants and products constant over time. This equilibrium is important for understanding how long a system can sustain production at a given rate without additional changes.

For ozone, the steady-state concentration can be studied through the balance of the involved reactions. Consider the reactions:
  • NO + O₃ → NO₂ + O₂ and
  • NO₂ + O₃ → NO + 2O₂.
In the steady state, the concentration of ozone is directly proportional to the concentration of NO₂ and inversely proportional to the concentration of NO. This means that for a stable level of ozone to be achieved, these substances must react at equal rates.

This understanding allows us to derive mathematical expressions that predict ozone concentrations based on current levels of NO and NO₂, thereby assisting in the control of air quality and pollution.
Rate Law Parameters
Rate law parameters are essential for quantifying how quickly chemical reactions occur. These parameters include the rate constants, which are specific for each reaction, and they dictate how concentrations of substances affect the reaction rate.

In the context of ozone formation, the rate laws for the reactions are written as:
  • For NO + O₃: \( r_1 = k_1 [NO][O_3] \),
  • And for NO₂ + O₃: \( r_2 = k_2 [NO_2][O_3] \).
Where \( k_1 \) and \( k_2 \) are the rate constants for reactions involving NO and NO₂, respectively.

The relationship between these parameters not only helps in predicting the rates at which ozone will form but also influences the decisions made in engineering processes to optimize them for minimal environmental impact. Properly understanding rate law parameters is vital for designing effective environmental protection measures.
Reaction Mechanisms
A reaction mechanism is a step-by-step sequence of elementary reactions by which an overall chemical change occurs. Understanding the precise mechanism of a reaction helps us comprehend the intricacies of how substances transform and allows us to design better catalytic systems to control unwanted reactions.

For ozone formation, understanding the steps involved helps in grasping how NO and NO₂ function within the overall photochemical cycle. Without NO and NO₂, ozone generation occurs via alternative mechanisms. Consider the following steps:
  • O₂ + M → 2O + M (oxygen dissociates)
  • O + O₂ + M → O₃ + M (ozone forms)
  • O₃ + M → O₂ + O + M (ozone dissociates)
These reactions highlight an autocatalytic behavior where molecules act as catalysts for their own reaction.

Understanding reaction mechanisms like these inform the development of models that predict how changes in conditions, like the absence of NO or NO₂, affect ozone levels. It can help in designing strategies to manage pollution levels more effectively.

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

The bacteria X-II can be described by a simple Monod equation with \(\mu_{\max }=$$$0.8 \mathrm{h}^{-1} \text {and } K_{\mathrm{M}}=4 . Y_{pc}=0.2 \mathrm{g} / \mathrm{g}, \text { and } Y_{\mathcal{W}}=2 \mathrm{g} / \mathrm{g} . \text { The process is carried out }$$ in a CSTR in which the feed rate is \)1000 \mathrm{dm}^{3} / \mathrm{h}\( at a substrate concentration of \)10 \mathrm{g} / \mathrm{dm}^{3}\( (a) What size fermentor is needed to achieve \)90 \%\( conversion of the strate? What is the exiting cell concentration? (b) How would your answer to (a) change if all the cells were filtered out and returned to the feed stream? (c) Consider now two \)5000 \mathrm{dm}^{3}\( CSTRs connect in series. What are the exiting concentrations \)C_{s}, C_{c},\( and \)C_{p}\( from each of the reactors? (d) Determine, if possible, the volumetric flow rate at which wash-out occurs and also the flow rate at which the cell production rate \)\left(C_{c} v_{0}\right)\( in grams per day is a maximum. (e) Suppose you could use the two \)5000-\mathrm{dm}^{3}\( reactors as batch reactors that take two hours to empty, clean, and fill. What would your production rate be in (grams per day) if your initial cell concentration is \)0.5 \mathrm{g} / \mathrm{dm}^{3} ?\( How many 500 -dm \)^{3}$ reactors would you need to match the CSTR production rate? (f) List ways you can work this problem incorrectly. (g) How could you make this problem more difficult?

(Postacidification in yogurt) Yogurt is produced by adding two strains of bacteria (Lactobacillus bulgaricus and Streptococcus thermophilus) to pasteurized milk. At temperatures of \(110^{\circ} \mathrm{F}\), the bacteria grow and produce lactic acid. The acid contributes flavor and causes the proteins to coagulate, giving the characteristic properties of yogurt. When sufficient acid has been produced (about \(0.90 \%\) ), the yogurt is cooled and stored until eaten by consumers. A lactic acid level of \(1.10 \%\) is the limit of acceptability. One limit on the shelf life of yogurt is "postacidification," or continued production of acid by the yogurt cultures during storage. The table that follows shows acid production (\% lactic acid) in yogurt versus time at four different temperatures. Acid production by yogurt cultures is a complex biochemical process. For the purpose of this problem, assume that acid production follows first-order kinetics with respect to the consumption of lactose in the yogurt to produce lactic acid. At the start of acid production the lactose concentration is about \(1.5 \%,\) the bacteria concentration is \(10^{11}\) cells/dm \(^{3}\), and the acid concentration at which all metabolic activity ceases is \(1.4 \%\) lactic acid. (a) Determine the activation energy for the reaction. (b) How long would it take to reach \(1.10 \%\) acid at \(38^{\circ} \mathrm{F}\) ? (c) If you left yogurt out at room temperature, \(77^{\circ} \mathrm{F}\), how long would it take to reach 1.10\% lactic acid? (d) Assuming that the lactic acid is produced in the stationary state, do the data fit any of the modules developed in this chapter? [Problem developed by General Mills, Minneapolis, Minnesota] (e) List ways you can work this problem incorrectly. (f) How could you make this problem more difficult?

The production of a product \(P\) from a particular gram negative bacteria follows the Monod growth law $$-r_{s}=\frac{\mu_{\max }^{\cdot} C_{s} C_{c}}{K_M+C_{s}}$$$$\text { with } \mu_{\max }=1 \mathrm{h}^{-1} K_{\mathrm{M}}=0.25 \mathrm{g} / \mathrm{dm}^{3}, \text { and } Y_{c / s}=0.5 \mathrm{g} / \mathrm{g}$$ (a) The reaction is to be carried out in a batch reactor with the initial cell concentration of \(C_{c 0}=0.1 \mathrm{g} / \mathrm{dm}^{3}\) and substrate concentration of \(C_{30}=20\) \(\mathrm{g} / \mathrm{dm}^{3}\) $$C_{c}=C_{c 0}+Y_{c, 4}\left(C_{s 0}-C_{s}\right)$$ Plot \(-r_{s}-r_{c}, C_{s},\) and \(C_{c}\) as a function of time. (b) Redo part (a) and use a logistic growth law. $$r_{x}=\mu_{\max }\left(1-\frac{C_{c}}{C_{x}}\right) c_{c}$$ and plot \(C_{c}\) and \(r_{c}\) as a function of time. The term \(C_{\infty}\) is the maximum cell mass and is called the carrying capacity and is equal to \(C_{\infty}=1.0\) \(\mathrm{g} / \mathrm{dm}^{3} .\) Can you find an analytical solution for the batch reactor? Compare with part (a) for \(C_{\infty}=Y_{c / s} C_{s 0}+C_{c 0}\) (c) The reaction is now to be carried out in a CSTR with \(C_{s0}=20 \mathrm{g} / \mathrm{dm}^{3}\) and \(C_{\mathrm{c} 0}=0 .\) What is the dilution rate at which washout occurs? (d) For the conditions in part (c), what is the dilution rate that will give the maximum product rate \((\mathrm{g} / \mathrm{h})\) if \(Y_{p / c}=0.15 \mathrm{g} / \mathrm{g} ?\) What are the concentrations \(C_{c}, C_{s}, C_{p},\) and \(-r_{s}\) at this value of \(D ?\) (e) How would your answers to (c) and (d) change if cell death could not be neglected with \(k_{d}=0.02 \mathrm{h}^{-1} ?\) (f) How would your answers to (c) and (d) change if maintenance could not be neglected with \(m=0.2 \mathrm{g} / \mathrm{h} / \mathrm{dm}^{3} ?\) (g) List ways you can work this problem incorrectly. (h) How could you make this problem more difficult?

(Open-ended problem) You may have to look up/guess/vary some of the constants. If methanol is ingested, it can be metabolized to formaldehyde, which can cause blindness if the formaldehyde reaches a concentration of 0.16 \(\mathrm{g} / \mathrm{dm}^{3}\) of fluid in the body. A concentration of \(0.75 \mathrm{g} / \mathrm{dm}^{3}\) will be lethal. After all the methanol has been removed from the stomach, the primary treatment is to inject ethanol intravenously to tie up (competitive inhibition) the enzyme alcohol dehydrogenase (ADH) so that methanol is not converted to formaldehyde and is eliminated from the body through the kidney and bladder \(\left(k_{7}\right) .\) We will assume as a first approximation that the body is a well-mixed CSTR of \(40 \mathrm{dm}^{3}\) (total body fluid). In Section \(7.5 .\) we applied a more rigorous model. The following reaction scheme can be applied to the body. The complete data set for this reaction is given on the CD-ROM, P7-25 and the open-ended problem H.10. After running the base case, vary the parameters and describe what you find.

The enzymatic hydrolysis of starch was carried out with and without maltose and \(\alpha\) -dextrin added. [Adapted from S. Aiba, A. E. Humphrey, and N.F. Mills, Biochemical Engineering (New York: Academic Press, 1973 ).

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.

Sign-up for free