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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

Mammalian cells can be cultured for a variety of purposes, including synthesis of vaccines. They must be maintained in growth media containing all of the components required for proper cellular function to ensure their survival and propagation. Traditionally, growth media were prepared by blending a powder, such as Dulbecco's Modified Eagle Medium (DMEM) with sterile deionized water. DMEM contains glucose, buffering agents, proteins, and amino acids. Using a sterile (i.e., bacterial-, fungal-,and yeast-free) growth medium ensures proper cell growth, but sometimes the water (or powder) can become contaminated, requiring the addition of antibiotics to eliminate undesired contaminants. The culture medium is supplemented with fetal bovine serum (FBS) that contains additional growth factors required by the cells. Suppose an aqueous stream (SG = 0.90) contaminated with bacteria is split, with 75\% being fed to a mixing unit to dissolve a powdered mixture of DMEM contaminated with the same bacteria found in the water. The ratio of impure feed water to powder entering the mixer is 4.4:1. The stream leaving the mixer (containing DMEM, water, and bacteria) is combined with the remaining 25\% of the aqueous stream and fed to a filtration unit to remove all of the bacteria that have contaminated the system, a total of \(20.0 \mathrm{kg}\). Once the bacteria have been removed, the sterile medium is combined with FBS and the antibiotic cocktail PSG (Penicillin-Streptomycin-L-Glutamine) in a shaking unit to generate 5000 L of growth medium (SG = 1.2). The final composition of the growth medium is 66.0 wt\% H_O, 11.0\% FBS, 8.0\% PSG, and the balance DMEM. (a) Draw and label the process flowchart. (b) Do a degree-of-freedom analysis around each piece of equipment (mixer, filter, and shaker), the splitter, the mixing point, and the overall system. Based on the analysis, identify which system or piece of equipment should be the starting point for further calculations. (c) Calculate all of the unknown process variables. (d) Determine a value for (i) the mass ratio of sterile growth medium product to feed water and (ii) the mass ratio of bacteria in the water to bacteria in the powder. (e) Suggest two reasons why the bacteria should be removed from the system.

The gas-phase reaction between methanol and acetic acid to form methyl acetate and water takes place in a batch reactor. When the reaction mixture comes to equilibrium, the mole fractions of the four reactive species are related by the reaction equilibrium constant $$K_{y}=\frac{y_{C} y_{D}}{y_{A} y_{B}}=4.87$$ (a) Suppose the feed to the reactor consists of \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0}, n_{\mathrm{D} 0},\) and \(n_{10}\) gram-moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and an inert gas, I, respectively. Let \(\xi\) be the extent of reaction. Write expressions for the gram-moles of each reactive species in the final product, \(n_{\mathrm{A}}(\xi), n_{\mathrm{B}}(\xi), n_{\mathrm{C}}(\xi),\) and \(n_{\mathrm{D}}(\xi) .\) Then use these expressions and the given equilibrium constant to derive an equation for \(\xi_{c}\), the equilibrium extent of reaction, in terms of \(\left.n_{\mathrm{A} 0}, \ldots, n_{10} . \text { (see Example } 4.6-2 .\right)\) (b) If the feed to the reactor contains equimolar quantities of methanol and acetic acid and no other species, calculate the equilibrium fractional conversion. (c) It is desired to produce 70 mol of methyl acetate starting with 75 mol of methanol. If the reaction proceeds to equilibrium, how much acetic acid must be fed? What is the composition of the final product? (d) Suppose it is important to reduce the concentration of methanol by making its conversion at equilibrium as high as possible, say 99\%. Again assuming the feed to the reactor contains only methanol and acetic acid and that it is desired to produce 70 mol of methyl acetate, determine the extent of reaction and quantities of methanol and acetic acid that must be fed to the reactor. (e) If you wanted to carry out the process of Part (b) or (c) commercially, what would you need to know besides the equilibrium composition to determine whether the process would be profitable? (List several things.)

An evaporation-crystallization process of the type described in Example \(4.5-2\) is used to obtain solid potassium sulfate from an aqueous solution of this salt. The fresh feed to the process contains 19.6 wt\% \(\mathrm{K}_{2} \mathrm{SO}_{4}\). The wet filter cake consists of solid \(\mathrm{K}_{2} \mathrm{SO}_{4}\) crystals and a \(40.0 \mathrm{wt} \% \mathrm{K}_{2} \mathrm{SO}_{4}\) solution, in a ratio \(10 \mathrm{kg}\) crystals/kg solution. The filtrate, also a \(40.0 \%\) solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0\% is evaporated. The evaporator has a maximum capacity of 175 kg water evaporated/s. (a) Assume the process is operating at maximum capacity. Draw and label a flowchart and do the degree-of-freedom analysis for the overall system, the recycle-fresh feed mixing point, the evaporator, and the crystallizer. Then write in an efficient order (minimizing simultaneous equations) the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don't do the calculations. (b) Calculate the maximum production rate of solid \(\mathrm{K}_{2} \mathrm{SO}_{4}\), the rate at which fresh feed must be supplied to achieve this production rate, and the ratio kg recycle/kg fresh feed. (c) Calculate the composition and feed rate of the stream entering the crystallizer if the process is scaled to 75\% of its maximum capacity. (d) The wet filter cake is subjected to another operation after leaving the filter. Suggest what it might be. Also, list what you think the principal operating costs for this process might be. (e) Use an equation-solving computer program to solve the equations derived in Part (a). Verify that you get the same solutions determined 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.

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?]
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