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Chapter 4: Problem 67

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carbon dioxide and water: $$\begin{aligned} \mathrm{CH}_{4}+\mathrm{O}_{2} & \rightarrow \mathrm{HCHO}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{CH}_{4}+2 \mathrm{O}_{2} & \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ The feed to the reactor contains equimolar amounts of methane and oxygen. Assume a basis of \(100 \mathrm{mol}\) feed/s. (a) Draw and label a flowchart. Use a degree-of-freedom analysis based on extents of reaction to determine how many process variable values must be specified for the remaining variable values to be calculated. (b) Use Equation 4.6-7 to derive expressions for the product stream component flow rates in terms of the two extents of reaction, \(\xi_{1}\) and \(\xi_{2}\) (c) The fractional conversion of methane is 0.900 and the fractional yield of formaldehyde is 0.855 . Calculate the molar composition of the reactor output stream and the selectivity of formaldehyde production relative to carbon dioxide production. (d) A classmate of yours makes the following observation: "If you add the stoichiometric equations for the two reactions, you get the balanced equation $$2 \mathrm{CH}_{4}+3 \mathrm{O}_{2} \rightarrow \mathrm{HCHO}+\mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}$$ The reactor output must therefore contain one mole of \(\mathrm{CO}_{2}\) for every mole of HCHO, so the selectivity of formaldehyde to carbon dioxide must be \(1.0 .\) Doing it the way the book said to do it, \(I\) got a different selectivity. Which way is right, and why is the other way wrong?" What is your response?

Short Answer

Expert verified
The molar composition of the reactor output is \(CH_4:0.0414\), \(HCHO:0.3556\), \(H_2O:0.1207\), \(CO_2:0.0603\), \(O_2:0.4220\). The selectivity of formaldehyde to carbon dioxide is 5.9. The selectivity would not be 1.0, as not all methane in the reactor forms CO2 and HCHO in a 1:1 ratio due to independent occurrence of reactions.

Step by step solution

01

Degree of freedom analysis

The degree of freedom is calculated using the formula \(F=C鈭扨+2\), where \(C\) is the number of components and \(P\) is the number of independent chemical reactions (or processes). We have 5 components (\(CH_4\), \(O_2\), \(HCHO\), \(H_2O\), and \(CO_2\)) and 2 independent reactions, so that results in: \(F=5鈭�2+2=5\). This means 5 process variable values must be specified.
02

Derive expressions for product stream component

Using the stoichiometric relationships in the chemical reactions and given feed molar rates, we can express product stream molar flow rates as follows: \(\dot{n}_{CH4} =100-\xi_1 - \xi_2\), \(\dot{n}_{O_2} =100-\xi_1 - 2 \xi_2\), \(\dot{n}_{HCHO} = \xi_1\), \(\dot{n}_{H_2O} =2 \xi_2\), and \(\dot{n}_{CO2} = \xi_2\). Here \(\xi_1\) and \(\xi_2\) represent the extents of the first and second reactions respectively.
03

Compute the molar composition of the reactor output stream

The conversion of methane is given as 0.900, and fractional yield of formaldehyde is 0.855. Therefore, we know that \(0.900 (100) = 90\) moles of methane react and that \(0.855 (90) = 76.95\) moles of formaldehyde are formed by reaction 1. This leads us to \(90 - 76.95 = 13.05\) moles of methane react according to reaction 2, producing \(13.05\) moles of CO2 and \(2*13.05 = 26.1\) moles of water. Total moles in output = \(100 +76.95 + 13.05 +26.1 = 216.1\) moles. The molar composition will be \(n_{CH4} = 0.0414\), \(n_{HCHO} = 0.3556\), \(n_{H2O} = 0.1207\), \(n_{CO2} = 0.0603\), \(n_{O2} = 0.4220\).
04

The selectivity of formaldehyde production

The selectivity is the molar flow rate of the desired product to that of the undesired product. Hence, the selectivity of HCHO to CO2 is \(\frac{n_{HCHO}}{n_{CO2}} = \frac{0.3556}{0.0603} = 5.9\).
05

Evaluate the classmate's observation

Your classmate's observation is incorrect. Although the equations can be algebraically added and simplify to his equation, the reactors in fact do not operate at overall stoichiometric conditions. Therefore, simply summing the reactions doesn't appropriately represent the individual influence of each reaction in the actual process. That's why the stoichiometric analysis on a per-reaction basis gives a different selectivity value, which is the correct one.

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Key Concepts

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

Stoichiometry
Stoichiometry in chemical reactions is essentially a recipe that tells us how much of each reactant is needed to produce a given amount of product. It's like following a recipe to make a cake, where specific amounts of ingredients are required. In the provided problem, we are looking at two reactions involving methane \( (\mathrm{CH}_4) \) and oxygen \( (\mathrm{O}_2) \). One forms formaldehyde \( (\mathrm{HCHO}) \) and water \( (\mathrm{H}_2O) \), while the other forms carbon dioxide \( (\mathrm{CO}_2) \) and water.

Understanding stoichiometry is crucial because it lets us calculate how much product we can get given certain amounts of reactants, provided we know the ratios given in the chemical equations.
- The first reaction suggests that one mole of methane reacts with one mole of oxygen to produce formaldehyde and water.
- The second reaction shows that one mole of methane reacts with two moles of oxygen to produce carbon dioxide and two moles of water.

Overall, stoichiometry helps us proportion the chemical "ingredients" in the correct ratios to achieve the desired reactions, as described in the problem's chemical equations.
Degree of Freedom Analysis
Degree of Freedom Analysis is a key concept in chemical engineering that helps determine how many variables in a system need to be specified before the remaining variables can be calculated. It's like solving a puzzle, where some pieces (variables) need to be in place before the rest can fit.

In our exercise, we use the degree of freedom formula \( F = C-P+2 \), where \( C \) is the number of components, \( P \) is the number of independent reactions.
- Here, we have 5 components: methane \( (\mathrm{CH}_4) \), oxygen \( (\mathrm{O}_2) \), formaldehyde \( (\mathrm{HCHO}) \), water \( (\mathrm{H}_2O) \), and carbon dioxide \( (\mathrm{CO}_2) \).
- There are 2 independent reactions happening.

Plugging these into the formula, we find \( F = 5 - 2 + 2 = 5 \). This means we need to find values for 5 processing variables to completely describe the system. These variables could be molar flow rates, extents of reaction, or other measurable quantities, each vital to solving the chemical reaction puzzle in the reactor.
Selectivity in Chemical Reactions
Selectivity in chemical reactions refers to how preferentially a particular reaction pathway is followed to produce the desired product over undesirable side-reactions. It's essentially a measure of efficiency, showing how much of a desired product is made compared to a waste product.

In our problem, selectivity is calculated for formaldehyde \( (\mathrm{HCHO}) \) relative to carbon dioxide \( (\mathrm{CO}_2) \). This tells us how effectively the process produces formaldehyde over CO2.

- The molar flow rate of formaldehyde \( n_{\mathrm{HCHO}} \) compared to carbon dioxide \( n_{\mathrm{CO}_2} \) gives us the selectivity ratio. In this problem, selectivity is calculated as \( \frac{n_{\mathrm{HCHO}}}{n_{\mathrm{CO}_2}} = \frac{0.3556}{0.0603} \), resulting in a selectivity of around 5.9.
- This means that for every mole of CO2 produced, approximately 5.9 moles of formaldehyde are produced.

Understanding selectivity helps improve industrial processes by optimizing conditions to maximize desired production and minimize waste, which is crucial for economic and environmental reasons.

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Most popular questions from this chapter

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?

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.)

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?]

In the production of a bean oil, beans containing 13.0 wt\% oil and \(87.0 \%\) solids are ground and fed to a stirred tank (the extractor) along with a recycled stream of liquid \(n\) -hexane. The feed ratio is \(3 \mathrm{kg}\) hexane/kg beans. The ground beans are suspended in the liquid, and essentially all of the oil in the beans is extracted into the hexane. The extractor effluent passes to a filter where the solids are collected and form a filter cake. The filter cake contains 75.0 wt\% bean solids and the balance bean oil and hexane, the latter two in the same ratio in which they emerge from the extractor. The filter cake is discarded and the liquid filtrate is fed to a heated evaporator in which the hexane is vaporized and the oil remains as a liquid. The oil is stored in drums and shipped. The hexane vapor is subsequently cooled and condensed, and the liquid hexane condensate is recycled to the extractor. (a) Draw and label a flowchart of the process, do the degree-of-freedom analysis, and write in an efficient order the equations you would solve to determine all unknown stream variables, circling the variables for which you would solve. (b) Calculate the yield of bean oil product (kg oil/kg beans fed), the required fresh hexane feed \(\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{14} / \mathrm{kg} \text { beans fed }\right),\) and the recycle to fresh feed ratio (kg hexane recycled/kg fresh feed). (c) It has been suggested that a heat exchanger might be added to the process. This process unit would consist of a bundle of parallel metal tubes contained in an outer shell. The liquid filtrate would pass from the filter through the inside of the tubes and then go on to the evaporator. The hot hexane vapor on its way from the evaporator to the extractor would flow through the shell, passing over the outside of the tubes and heating the filtrate. How might the inclusion of this unit lead to a reduction in the operating cost of the process? (d) Suggest additional steps that might improve the process economics.

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.
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