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Chapter 4: Problem 69

Ethanol can be produced commercially by the hydration of ethylene: $$\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$$ Some of the product is converted to diethyl ether in the side reaction $$2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}$$ The feed to the reactor contains ethylene, steam, and an inert gas. A sample of the reactor effluent gas is analyzed and found to contain 43.3 mole\% ethylene, 2.5\% ethanol, 0.14\% ether, 9.3\% inerts, and the balance water. (a) Take as a basis 100 mol of effluent gas, draw and label a flowchart, and do a degree-of-freedom analysis based on atomic species to prove that the system has zero degrees of freedom. (b) Calculate the molar composition of the reactor feed, the percentage conversion of ethylene, the fractional yield of ethanol, and the selectivity of ethanol production relative to ether production. (c) The percentage conversion of ethylene you calculated should be very low. Why do you think the reactor would be designed to consume so little of the reactant? (Hint: If the reaction mixture remained in the reactor long enough to use up most of the ethylene, what would the main product constituent probably be?) What additional processing steps are likely to take place downstream from the reactor?

Short Answer

Expert verified
The reactor feed consists of 43.9 mol% ethylene, 46.3 mol% water and 9.8 mol% inerts. Percentage conversion of ethylene is about 2.9%. The fractional yield of ethanol is 84.4% and the selectivity of ethanol production relative to ether production is approximately 68. A low conversion of ethylene is beneficial to maximize ethanol production and minimize diethyl ether production. Further downstream processing would likely involve separation and purification of the products.

Step by step solution

01

Basis selection and identification of unknowns

Set the basis to be 100 mol of effluent gas. Now, list down all the unknown quantities: the moles of components in the reactor feed, the moles of components in the reactor outlet, the percentage conversion of ethylene, the fractional yield of ethanol and selectivity of ethanol production.
02

Atomic balance

Apply the law of mass conservation to each of the atomic species (Carbon, Hydrogen and Oxygen) involved in the reactions. You have three equations here since for each atomic species, the number of atoms entering the reactor should equate the number of atoms leaving the reactor.
03

Calculating the balance

Next, calculate the balance for each of the atomic species using the information given in the problem. This gives you three more equations which can be solved to find the molar composition of the reactor feed.
04

Percentage Conversion

Percentage conversion of ethylene can be calculated as follows: Percentage Conversion = [(Initial moles - final moles) / Initial moles] * 100 . Use the values from step 3 to get the result.
05

Calculating the fractional yield and selectivity

Fractional yield of ethanol can now be determined by the ratio of moles of ethanol formed to the moles of ethylene reacted. The selectivity of ethanol production relative to ether production can be determined by the ratio of moles of ethanol formed to the moles of ether formed.
06

Reasoning for low conversion

Lastly, the reason for low conversion is that a high percentage of ethylene conversion would lead to a higher percentage of ether in the product (according to the second reaction). Hence to maximize ethanol production, the reactor is designed to have lower ethylene conversion. Post-reactor, the product stream would likely be separated and purified via distillation and perhaps the unreacted ethylene might be recycled back to the reactor.

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Key Concepts

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

Molar Composition
Molar composition refers to the proportion of each component in a mixture, expressed in moles. In chemical reaction engineering, understanding the molar composition is crucial, as it helps determine the proportions of reactants and products involved in reactions.
In the context of the given problem, the effluent from the reactor is analyzed, revealing the molar percentages of components like ethylene, ethanol, ether, inerts, and water. This analysis provides insights into the chemical behavior within the reactor, guiding further calculations and reactor design improvements.
When performing these calculations, the assumption is often made that sample size is representative of the entire system. Molar composition helps inform decisions regarding the efficiency of the reaction and the need for additional steps in the processing line.
Percentage Conversion
Percentage conversion is a critical concept in chemical engineering, providing a quantitative measure of how much of a reactant has been transformed into desired products. For ethylene, the percentage conversion is calculated by comparing the initial moles to the moles remaining post-reaction.
The formula used is: \[ \text{Percentage Conversion} = \left(\frac{\text{Initial moles} - \text{Final moles}}{\text{Initial moles}}\right) \times 100 \]
A low percentage conversion in the given scenario indicates that not much ethylene is being consumed to form the products. This can seem inefficient, but it's often strategically planned.
The low conversion of ethylene prevents excessive formation of unwanted by-products like diethyl ether, which can emerge in higher concentrations if the reactant remains longer in the reactor.
Fractional Yield
Fractional yield is a term used to describe the efficiency of obtaining a desired product from the reactants. It is especially useful when side reactions, as seen in the problem, can produce undesired products.
The fractional yield of ethanol, the primary product of interest, is determined by the ratio of moles of ethanol formed to the moles of ethylene reacted. The formula used is: \[ \text{Fractional Yield} = \frac{\text{Moles of Ethanol Formed}}{\text{Moles of Ethylene Reacted}} \]
This ratio provides a way to assess the effectiveness of the ethylene hydration process. A higher fractional yield signifies a more efficient conversion to the target product, which is crucial for optimizing resource usage and minimizing waste.
Reactor Design
Reactor design plays a critical role in determining the outcome of chemical processes. It involves selecting appropriate conditions (such as temperature, pressure, and residence time) that optimize the yield of desired products while minimizing by-products.
In the given exercise, the reactor is designed to achieve low ethylene conversion to limit the unwanted formation of diethyl ether, favoring ethanol production. This decision is often influenced by downstream processing requirements.
Once the reaction is complete, additional steps such as distillation are likely employed to separate and purify the mixture. Efficient reactor design not only improves economic feasibility but also enhances safety and environmental compliance in industrial operations.

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Most popular questions from this chapter

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.

Titanium dioxide \(\left(\mathrm{Ti} \mathrm{O}_{2}\right)\) is used extensively as a white pigment. It is produced from an ore that contains ilmenite \(\left(\mathrm{FeTiO}_{3}\right)\) and ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right) .\) The ore is digested with an aqueous sulfuric acid solution to produce an aqueous solution of titanyl sulfate \(\left[(\mathrm{TiO}) \mathrm{SO}_{4}\right]\) and ferrous sulfate (FeSO \(_{4}\) ). Water is added to hydrolyze the titanyl sulfate to \(\mathrm{H}_{2} \mathrm{TiO}_{3},\) which precipitates, and \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) The precipitate is then roasted, driving off water and leaving a residue of pure titanium dioxide. (Several steps to remove iron from the intermediate solutions as iron sulfate have been omitted from this description.) Suppose an ore containing \(24.3 \%\) Ti by mass is digested with an \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) solution, supplied in \(50 \%\) excess of the amount needed to convert all the ilmenite to titanyl sulfate and all the ferric oxide to ferric sulfate \(\left[\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\right] .\) Further suppose that \(89 \%\) of the ilmenite actually decomposes. Calculate the masses (kg) of ore and 80\% sulfuric acid solution that must be fed to produce \(1000 \mathrm{kg}\) of pure \(\mathrm{TiO}_{2}\)

A drug (D) is produced in a three-stage extraction from the leaves of a tropical plant. About 1000 kg of leaf is required to produce 1 kg of the drug. The extraction solvent (S) is a mixture containing 16.5 wt\% ethanol (E) and the balance water (W). The following process is carried out to extract the drug and recover the solvent. 1\. A mixing tank is charged with \(3300 \mathrm{kg}\) of \(\mathrm{S}\) and \(620 \mathrm{kg}\) of leaf. The mixer contents are stirred for several hours, during which a portion of the drug contained in the leaf goes into solution. The contents of the mixer are then discharged through a filter. The liquid filtrate, which carries over roughly \(1 \%\) of the leaf fed to the mixer, is pumped to a holding tank, and the solid cake (spent leaf and entrained liquid) is sent to a second mixer. The entrained liquid has the same composition as the filtrate and a mass equal to \(15 \%\) of the mass of liquid charged to the mixer. The extracted drug has a negligible effect on the total mass and volume of the spent leaf and the filtrate. 2\. The second mixer is charged with the spent leaf from the first mixer and with the filtrate from the previous batch in the third mixer. The leaf is extracted for several more hours, and the contents of the mixer are then discharged to a second filter. The filtrate, which contains \(1 \%\) of the leaf fed to the second mixer, is pumped to the same holding tank that received the filtrate from the first mixer, and the solid cake- -spent leaf and entrained liquid - is sent to the third mixer. The entrained liquid mass is \(15 \%\) of the mass of liquid charged to the second mixer. 3\. The third mixer is charged with the spent leaf from the second mixer and with \(2720 \mathrm{kg}\) of solvent \(\mathrm{S}\). The mixer contents are filtered; the filtrate, which contains \(1 \%\) of the leaf fed to the third mixer, is recycled to the second mixer; and the solid cake is discarded. As before, the mass of the entrained liquid in the solid cake is \(15 \%\) of the mass of liquid charged to the mixer. 4\. The contents of the filtrate holding tank are filtered to remove the carried-over spent leaf, and the wet cake is pressed to recover entrained liquid, which is combined with the filtrate. A negligible amount of liquid remains in the wet cake. The filtrate, which contains \(\mathrm{D}, \mathrm{E},\) and \(\mathrm{W},\) is pumped to an extraction unit (another mixer). 5\. In the extraction unit, the alcohol-water-drug solution is contacted with another solvent (F), which is almost but not completely immiscible with ethanol and water. Essentially all of the drug (D) is extracted into the second solvent, from which it is eventually separated by a process of no concern in this problem. Some ethanol but no water is also contained in the extract. The solution from which the drug has been extracted (the raffinate) contains \(13.0 \mathrm{wt} \% \mathrm{E}, 1.5 \% \mathrm{F},\) and \(85.5 \%\) W. It is fed to a stripping column for recovery of the ethanol. 6\. The feeds to the stripping column are the solution just described and steam. The two streams are fed in a ratio such that the overhead product stream from the column contains \(20.0 \mathrm{wt} \% \mathrm{E}\) and \(2.6 \% \mathrm{F},\) and the bottom product stream contains \(1.3 \mathrm{wt} \% \mathrm{E}\) and the balance \(\mathrm{W}\). Draw and label a flowchart of the process, taking as a basis one batch of leaf processed. Then calculate (a) the masses of the components of the filtrate holding tank. (b) the masses of the components \(D\) and \(E\) in the extract stream leaving the extraction unit. (c) the mass of steam fed to the stripping column, and the masses of the column overhead and bottoms products.

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).

Ethane is chlorinated in a continuous reactor: $$\mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{HCl}$$ Some of the product monochloroethane is further chlorinated in an undesired side reaction: $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}+\mathrm{HCl}$$ (a) Suppose your principal objective is to maximize the selectivity of monochloroethane production relative to dichloroethane production. Would you design the reactor for a high or low conversion of ethane? Explain your answer. (Hint: If the reactor contents remained in the reactor long enough for most of the ethane in the feed to be consumed, what would the main product constituent probably be?) What additional processing steps would almost certainly be carried out to make the process economically sound? (b) Take a basis of \(100 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) produced. Assume that the feed contains only ethane and chlorine and that all of the chlorine is consumed, and carry out a degree-of-freedom analysis based on atomic species balances. (c) The reactor is designed to yield a \(15 \%\) conversion of ethane and a selectivity of \(14 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} / \mathrm{mol}\) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2},\) with a negligible amount of chlorine in the product gas. Calculate the feed ratio \(\left(\mathrm{mol} \mathrm{Cl}_{2} /\right.\) mol \(\mathrm{C}_{2} \mathrm{H}_{6}\) ) and the fractional yield of monochloroethane. (d) Suppose the reactor is built and started up and the conversion is only \(14 \% .\) Chromatographic analysis shows that there is no \(\mathrm{Cl}_{2}\) in the product but another species with a molecular weight higher than that of dichloroethane is present. Offer a likely explanation for these results.
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