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Chapter 9: Problem 24

Synthetically produced ethanol is an important industrial commodity used for various purposes, including as a solvent (especially for substances intended for human contact or consumption); in coatings, inks, and personal-care products; for sterilization; and as a fuel. Industrial cthanol is a petrochemical synthesized by the hydrolysis of ethylene: $$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{v})$$ Some of the product is converted to diethyl ether in the undesired side reaction $$2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{v}) \rightleftharpoons\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}(\mathrm{v})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) $$The combined feed to the reactor contains 53.7 mole \(\% \mathrm{C}_{2} \mathrm{H}_{4}, 36.7 \% \mathrm{H}_{2} \mathrm{O}\) and the balance nitrogen, and enters the reactor at \(310^{\circ} \mathrm{C}\). The reactor operates isothermally at \(310^{\circ} \mathrm{C}\). An ethylene conversion of \(5 \%\) is achieved, and the yield of ethanol (moles cthanol produced/mole cthylene consumed) is 0.900 . Data for Diethyl Ether $$\begin{aligned}&\Delta \hat{H}_{f}^{\circ}=-271.2 \mathrm{kJ} / \mathrm{mol} \text { for the liquid }\\\ &\left.\Delta \hat{H}_{v}=26.05 \mathrm{kJ} / \mathrm{mol} \quad \text { (assume independent of } T\right)\end{aligned}$$ $$C_{p}\left[\mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]=0.08945+40.33 \times 10^{-5} T\left(^{\circ} \mathrm{C}\right)-2.244 \times 10^{-7} T^{2}$$ (a) Calculate the reactor heating or cooling requirement in \(\mathrm{kJ} / \mathrm{mol}\) feed. (b) Why would the reactor be designed to yield such a low conversion of ethylene? What processing step (or steps) would probably follow the reactor in a commercial implementation of this process?

Short Answer

Expert verified
(a) The reactor heating or cooling requirement should be calculated according to the steps above, considering the enthalpies of the main and side reactions and the stoichiometry of each reaction. (b) The reactor may be designed to yield such a low conversion of ethylene to limit the production of undesired side product and manage safety or financial concerns. A commercial process could likely include a purification step after the reactor to isolate the desired ethanol product.

Step by step solution

01

Determining chemical reactions

From the problem, there are two reactions taking place: The main reaction is from ethylene and water to produce ethanol \(C_{2}H_{4}(g) + H_{2}O(v) \rightarrow C_{2}H_{5}OH(v)\) and the side reaction is from ethanol to produce diethyl ether \(2C_{2}H_{5}OH(v) \rightarrow {(C_{2}H_{5})}_{2}O(v) + H_{2}O(v)\). For every mole of ethylene consumed, 0.900 moles of ethanol are produced, the remaining 0.100 moles must have reacted to produce diethyl ether.
02

Calculating the enthalpy of reactions

We know that for every mole ethylene used, 0.900 moles of ethanol are produced and 0.100 moles of diethyl ether are produced. Thus, using the given enthalpy data, we can figure out the total enthalpy of reactions. The enthalpy of reaction for the main reaction can be estimated using standard enthalpies of formation from stable forms. The same can be done for the side reaction. As an important detail, care should be taken to use the enthalpy of vaporization for diethyl ether in the side reaction.
03

Calculating the reactor heat

Using the calculated enthalpies of reactions for two reactions, we calculate the total heat or enthalpy change in the reactor per mole of ethylene consumed. As 5% of the incoming ethylene is consumed, we can extrapolate this value to kJ/mol of feed. An important point is to take into consideration that the process is performed at constant temperature, so any heat produced or consumed in the reaction will either have to be added (in case the reaction is endothermic) or removed (if the reaction is exothermic) to keep the temperature steady.
04

Understanding low conversion in a reactor

This is more of a theoretical question, but it essentially brings up concepts of product selectivity and avoiding unwanted side reactions. In this case, a high yield of unnecessary products in unfavorable conditions may be the reason for such a low conversion rate. A probable further step in commercial process may be purification, to extract the required ethanol product while minimizing the side product caused by the side reaction.

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Key Concepts

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

Enthalpy of Reaction
Enthalpy of reaction, often represented as \( \Delta H_{\text{reaction}} \) in chemistry, is a crucial concept when discussing chemical reactions. It defines the total heat content change during a chemical process. When we examine the synthesis of ethanol through the hydrolysis of ethylene, understanding the enthalpy of the main reaction and side reactions is essential.

The enthalpy of the desired ethanol-producing reaction and the undesired diethyl ether-producing side reaction are determined using standard heats of formation and, when applicable, enthalpy of vaporization. These values are critical for engineers to calculate the total energy change within the reactor, which will indicate whether the reactor will require heating or cooling to maintain its isothermal condition. If the reactions are endothermic (absorbing heat), energy must be supplied. Conversely, if they are exothermic (releasing heat), energy must be dissipated.

To accurately calculate the enthalpy of reactions in the synthesis of ethanol, one must take into account not only the standard conditions but also the specific conditions of the reactor, such as temperature and pressure, to make necessary adjustments to the standard enthalpy values.
Reactor Design
In the context of synthesizing industrial ethanol, reactor design plays a crucial role in achieving the desired conversion and selectivity of the desired product. The particular low ethylene conversion of 5% mentioned in the exercise raises the question of why the reactor is designed this way.

A key reason is the trade-off between conversion and selectivity. High conversions often drive the reaction towards more side products, like diethyl ether in this case, which are not desired. A reactor designed to favor higher selectivity towards ethanol at the cost of conversion might be preferable. Additionally, low conversion reactors are often set up in a series to gradually achieve the desired overall conversion while maintaining high selectivity.

Following the chemical reaction, further processing steps such as distillation or separation are common. These processes are designed to purify the ethanol from other substances, like diethyl ether and water. This reactor design approach reflects a careful balance between efficiency, product quality, and overall cost of the process.
Chemical Process Optimization
Chemical process optimization is about improving chemical reaction processes to achieve a set of objectives, such as increasing yield, improving product quality, and reducing energy consumption. It combines principles of chemistry, engineering, and economics.

For the industrial production of ethanol, optimization could focus on adjusting the proportion of reactants, improving catalyst effectiveness, or tweaking reaction conditions, such as temperature and pressure. Optimization considerations include reducing the formation of by-products like diethyl ether, conserving energy, and optimizing the use of feed materials.

Engineers and chemists must work in tandem to adjust variables, potentially using computational models to predict how changes will affect the process before physically implementing them. The exercise of calculating reactor heating or cooling requirements represents a basic form of optimization aimed at maintaining energy efficiency during the isothermal operation of the reactor.

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Most popular questions from this chapter

In a coal gasification process, carbon (the primary constituent of coal) reacts with steam to produce carbon monoxide and hydrogen (synthesis gas). The gas may either be burned or subjected to further processing to produce any of a variety of chemicals. A coal contains 10.5 wt\% moisture (water) and 22.6 wt\% noncombustible ash. The remaining fraction of the coal contains 81.2 wife \(\mathrm{C}, 13.4 \%\) O, and \(5.4 \%\) H. A coal slurry containing \(2.00 \mathrm{kg}\) coal/kg water is fed at \(25^{\circ} \mathrm{C}\) to an adiabatic gasification reactor along with a stream of pure oxygen at the same temperature. The following reactions take place in the reactor: $$\begin{array}{l}\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+131.3 \mathrm{kJ} \\\\\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-393.5 \mathrm{kJ} \\ 2 \mathrm{H}(\mathrm{in} \mathrm{coal})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ} \approx-242 \mathrm{kJ}\end{array}$$ Gas and slag (molten ash) leave the reactor at \(2500^{\circ} \mathrm{C}\). The gas contains \(\mathrm{CO}, \mathrm{H}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}^{14}\) (a) Feeding oxygen to the reactor lowers the yield of synthesis gas, but no gasifier ever operates without supplementary oxygen. Why does the oxygen lower the yield? Why it is nevertheless always supplied. (Hint: All the necessary information is contained in the first two stoichiometric equations and associated heats of reaction shown above.) (b) Suppose the oxygen gas fed to the reactor and the oxygen in the coal combine with all the hydrogen in the coal (Reaction 3) and with some of the carbon (Reaction 2), and the remainder of the carbon is consumed in Reaction 1. Taking a basis of 1.00 kg coal fed to the reactor and letting \(n_{0}\) equal the moles of \(\mathrm{O}_{2}\) fed, draw and label a flowchart. Then derive expressions for the molar flow rates of the four outlet gas species in terms of \(n_{0}\). [Partial solution: \(n_{\mathrm{H}_{2}}=\left(51.3-n_{0}\right)\) mol \(\mathrm{H}_{2} . \mathrm{J}\) (c) The standard heat of combustion of the coal has been determined to be -21,400 kJ/kg, taking \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) to be the combustion products. Use this value and the given clemental composition of the coal to prove that the standard heat of formation of the coal is \(-1510 \mathrm{kJ} / \mathrm{kg}\). Then use an energy balance to calculate \(n_{0},\) using the following approximate heat capacities in your calculation: Take the heat of fusion of ash (the heat required to convert ash to slag) to be \(710 \mathrm{kJ} / \mathrm{kg}\).

Methane at \(25^{\circ} \mathrm{C}\) is burned in a boiler furnace with \(10.0 \%\) excess air preheated to \(100^{\circ} \mathrm{C}\). Ninety percent of the methane fed is consumed, the product gas contains \(10.0 \mathrm{mol} \mathrm{CO}_{2} / \mathrm{mol} \mathrm{CO},\) and the combustion products leave the furnace at \(400^{\circ} \mathrm{C}\). (a) Calculate the heat transferred from the furnace, \(-\dot{Q}(\mathrm{kW}),\) for a basis of \(100 \mathrm{mol} \mathrm{CH}_{4}\) fed/s. (The greater the value of \(-\dot{Q}\), the more steam is produced in the boiler.) (b) Would the following changes increase or decrease the rate of steam production? (Assume the fuel feed rate and fractional conversion of methane remain constant.) Briefly explain your answers. (i) Increasing the temperature of the inlet air; (ii) increasing the percent excess air for a given stack gas temperature; (iii) increasing the selcctivity of \(\mathrm{CO}_{2}\) to \(\mathrm{CO}\) formation in the furnace; and (iv) increasing the stack gas temperature.

Lime (calcium oxide) is widely used in the production of cement, steel, medicines, insecticides, plant and animal food, soap, rubber, and many other familiar materials. It is usually produced by heating and decomposing limestone (CaCO \(_{3}\) ), a cheap and abundant mineral, in a calcination process: $$\mathrm{CaCO}_{3}(\mathrm{s}) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$$ (a) Limestone at \(25^{\circ} \mathrm{C}\) is fed to a continuous calcination reactor. The calcination is complete, and the products leave at \(900^{\circ} \mathrm{C}\). Taking 1 metric ton \((1000 \mathrm{kg})\) of limestone as a basis and clemental species \(\left[\mathrm{Ca}(\mathrm{s}), \mathrm{C}(\mathrm{s}), \mathrm{O}_{2}(\mathrm{g})\right]\) at \(25^{\circ} \mathrm{C}\) as references for enthalpy calculations, prepare and fill in an inlet-outlet enthalpy table and prove that the required heat transfer to the reactor is \(2.7 \times 10^{6} \mathrm{kJ}\) (b) In a common variation of this process, hot combustion gases containing oxygen and carbon monoxide (among other components) are fed into the calcination reactor along with the limestone. The carbon monoxide is oxidized in the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})$$ Suppose the combustion gas fed to a calcination reactor contains 75 mole \(\% \mathrm{N}_{2}, 2.0 \% \mathrm{O}_{2}, 9.0 \% \mathrm{CO},\) and \(14 \% \mathrm{CO}_{2}\) the gas enters the reactor at \(900^{\circ} \mathrm{C}\) in a feed ratio of \(20 \mathrm{kmol}\) gas/kmol limestone; the calcination is complete; all of the oxygen in the gas feed is consumed in the CO oxidation reaction; the reactor effluents are at \(900^{\circ} \mathrm{C}\) Again taking a basis of 1 metric ton of limestone calcined, prepare and fill in an inlet-outlet enthalpy table for this process [don't recalculate enthalpies already calculated in Part (a)] and calculate the required heat transfer to the reactor. (c) You should have found that the heat that must be transferred to the reactor is significantly lower with the combustion gas in the feed than it is without the gas. By what percentage is the heat requirement reduced? Give two reasons for the reduction. State another benefit of feeding the combustion gas, besides the reduction of the heating requirement.

An ultimate analysis of a coal is a series of operations that yields the percentages by mass of carbon, hydrogen, nitrogen, oxygen, and sulfur in the coal. The heating value of a coal is best determined in a calorimeter, but it may be estimated with reasonable accuracy from the ultimate analysis using the Dulong formula: $$H H V(\mathrm{k} J / \mathrm{kg})=33,801(\mathrm{C})+144,158[(\mathrm{H})-0.125(\mathrm{O})]+9413(\mathrm{S})$$ where (C), (H), (O), and (S) are the mass fractions of the corresponding elements. The 0.125(O) term accounts for the hydrogen bound in the water contained in the coal. (a) Derive an expression for the higher heating value ( \(H H V\) ) of a coal in terms of \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{S},\) and compare your result with the Dulong formula. Suggest a reason for the difference. (b) A coal with an ultimate analysis of \(75.8 \mathrm{wt} \% \mathrm{C}, 5.1 \% \mathrm{H}, 8.2 \% \mathrm{O}, 1.5 \% \mathrm{N}, 1.6 \% \mathrm{S},\) and \(7.8 \%\) ash (noncombustible) is burned in a power-plant boiler fumace. All of the sulfur in the coal forms \(\mathrm{SO}_{2}\) The gas leaving the furnace is fed through a tall stack and discharged to the atmosphere. The ratio \(\phi\) (\(\mathrm{kg} \mathrm{SO}_{2}\) in the stack gas/kJ heating value of the fuel) must be below a specified value for the power plant to be in compliance with Environmental Protection Agency regulations regarding sulfur emissions. Estimate \(\phi\), using the Dulong formula for the heating value of the coal. (c) An earlier version of the EPA regulation specified that the mole fraction of \(\mathrm{SO}_{2}\) in the stack gas must be less than a specified amount to avoid a costly fine and the required installation of an expensive stack gas scrubbing unit. When this regulation was in force, a few unethical plant operators blew clear air into the base of the stack while the furnace was operating. Briefly explain why they did so and why they stopped this practice when the new regulation was introduced.

In the production of many microelectronic devices, continuous chemical vapor deposition (CVD) processes are used to deposit thin and exceptionally uniform silicon dioxide films on silicon wafers. One CVD process involves the reaction between silane and oxygen at a very low pressure. $$\mathrm{SiH}_{4}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SiO}_{2}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g})$$ The feed gas, which contains oxygen and silane in a ratio \(8.00 \mathrm{mol} \mathrm{O}_{2} / \mathrm{mol} \mathrm{SiH}_{4},\) enters the reactor at 298 \(\mathrm{K}\) and 3.00 torr absolute. The reaction products emerge at \(1375 \mathrm{K}\) and 3.00 torr absolute. Essentially all of the silane in the feed is consumed. (a) Taking a basis of \(1 \mathrm{m}^{3}\) of feed gas, calculate the moles of each component of the feed and product mixtures and the extent of reaction, \(\xi\) (b) Calculate the standard heat of the silane oxidation reaction (kJ). Then, taking the feed and product species at \(298 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) as references, prepare an inlet-outlet enthalpy table and calculate and fill in the component amounts (mol) and specific enthalpies (kJ/mol). (See Example 9.5-1.) Data $$\left(\Delta \hat{H}_{\mathrm{f}}\right)_{\mathrm{SiH}_{4}(\mathrm{g})}=-61.9 \mathrm{kJ} / \mathrm{mol}, \quad\left(\Delta \hat{H}_{\mathrm{f}}^{\mathrm{o}}\right)_{\mathrm{SiO}_{2}(\mathrm{s})}=-851 \mathrm{kJ} / \mathrm{mol}$$ $$\left(C_{p}\right)_{\mathrm{SiH}_{4}(g)}[\mathrm{k} \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})]=0.01118+12.2 \times 10^{-5} T-5.548 \times 10^{-8} T^{2}+6.84 \times 10^{-12} T^{3}$$ $$\left(C_{p}\right)_{\mathrm{SiO}_{2}(\mathrm{s})}[\mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K})]=0.04548+3.646 \times 10^{-5} T-1.009 \times 10^{3} / T^{2}$$ The temperatures in the formulas for \(C_{p}\) are in kelvins. (c) Calculate the heat ( \(k\) J) that must be transferred to or from the reactor (state which it is). Then determine the required heat transfer rate ( \(\mathrm{kW}\) ) required for a reactor feed of \(27.5 \mathrm{m}^{3} / \mathrm{h}\).
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