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One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4-dicyanobutane. The reduction reaction is $$2 \mathrm{CH}_{2}=\mathrm{CHCN}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}$$ The \(\mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) is then chemically reduced using hydrogen gas to \(\mathrm{H}_{2} \mathrm{N}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2},\) which is used to produce production of nylon. What current must be used to produce \(150 . \mathrm{kg} \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) per hour?

Short Answer

Expert verified
The current required to produce 150 kg of NC-(CH2)4-CN per hour is approximately 77,263 A.

Step by step solution

01

Calculate moles of NC-(CH2)4-CN

First, we need to find the molar mass of NC-(CH2)4-CN. This can be done by adding the molar masses of each constituent element: N: 14 g/mol, C: 12 g/mol, H: 1 g/mol Molar mass of NC-(CH2)4-CN = 2 * 14 + 12 * 6 + 1 * 8 = 104 g/mol Now, we can find the moles of NC-(CH2)4-CN in 150 kg (or 150,000 g): Moles of NC-(CH2)4-CN = (150,000 g) / (104 g/mol) = 1442.31 mol
02

Calculate moles of electrons needed

Based on the balanced reduction reaction, we can see that 2 moles of electrons are required for 1 mole of NC-(CH2)4-CN. Therefore: Moles of electrons = 2 * Moles of NC-(CH2)4-CN = 2 * 1442.31 mol = 2884.62 mol
03

Calculate the total charge

One mole of electrons has a charge of Faraday's constant (F), which is approximately 96485 Coulombs/mol. Therefore: Total charge = Moles of electrons * F = 2884.62 mol * 96485 C/mol ≈ 278144724.7 C
04

Calculate the current

The problem asks for the current needed to produce 150 kg of NC-(CH2)4-CN per hour. Since current is defined as charge per unit time (I = Q/t), and we want the time to be one hour: Current = Total charge / Time = 278144724.7 C / (1 hour * 3600 s/hour) ≈ 77262.4 A Thus, at a current of approximately 77,263 A (rounded to the nearest integer), 150 kg of NC-(CH2)4-CN can be produced per hour.

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Key Concepts

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

Organic Synthesis
Organic synthesis is a fascinating aspect of chemistry, where complex organic compounds are constructed from simpler substances. In the context of the problem we're looking at, organic synthesis is pivotal in transforming starting materials into desired products like 1,4-dicyanobutane. This compound is a result of the reduction of 2CH extsubscript{2}=CHCN in the presence of protons and electrons. By breaking and forming specific chemical bonds in a controlled manner, organic synthesis allows us to prepare molecules which are crucial for various applications, including further steps in chemical processing. The significance of organic synthesis extends to myriad fields, such as pharmaceuticals, materials science, and agriculture. In our example, the outcome of synthetic steps serves as a precursor for further transformations essential for producing synthetic materials like nylon, demonstrating how organic synthesis serves as a cornerstone in creating innovative and useful products.
Industrial Chemistry
Industrial chemistry involves scaling up chemical reactions for practical applications in manufacturing. The production of 1,4-dicyanobutane is a classic example within this realm. It illustrates how industrial processes adapt laboratory-scale reactions to produce considerable quantities of chemicals to meet industrial needs. The Monsanto Company's use of electrochemistry to synthesize this compound reflects a broader trend where efficiency, scalability, and sustainability are crucial. Key industrial techniques focus on optimizing conditions such as temperature, pressure, and catalysis to maximize yield and purity. This often requires specialized equipment and systems to manage the reaction environment effectively. These principles not only optimize production but also ensure that the processes are economically viable and environmentally friendly. By understanding the mechanisms and practicalities of industrial chemistry, companies can deliver essential chemical products, such as those used in producing nylon, in large volumes necessary for global supply chains.
Current Calculation
In electrochemistry, calculating the required electrical current helps determine how much charge is needed to drive a chemical reaction. This concept is crucial for designing electrochemical cells and processes effectively, especially on an industrial scale like in the Monsanto process outlined. Current calculation hinges on Faraday’s laws of electrolysis, which relate the amount of substance transformed to the electric charge passed through the electrolyte. In this exercise, determining the current for producing 150 kg of 1,4-dicyanobutane involves several steps:
  • First, compute the moles of the compound based on its mass and molar mass.
  • Then, calculate the moles of electrons needed from the balanced chemical equation.
  • Next, find the total charge using Faraday’s constant.
  • Finally, divide this charge by the time (in seconds) to find the current.
This detailed calculation ensures that the desired amount of product can be produced reliably within a specified timeframe, here being one hour, by using a specified current of approximately 77,263 A.
Nylon Production
The process of nylon production underscores the importance of the industrial synthesis of 1,4-dicyanobutane. After forming this compound via electrochemical reduction, it is converted to hexamethylenediamine, a key precursor in making nylon polymers. Nylon, a kind of polyamide, is renowned for its strength, elasticity, and resistance to abrasion. These characteristics make it an ideal material for a wide array of products, from clothing and carpets to industrial threads and components. The production starts with the polymerization of hexamethylenediamine and adipic acid, forming long chains of repeating units. In this way, the electrochemical processes not only provide the starting materials for nylon but also highlight the interconnectedness of different chemistry branches. By building compounds meticulously, chemists can produce materials that significantly impact everyday life, bridging the gap between simple chemical reactions and sophisticated consumer products.

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Most popular questions from this chapter

The black silver sulfide discoloration of silverware can be removed by heating the silver article in a sodium carbonate solution in an aluminum pan. The reaction is $$3 \mathrm{Ag}_{2} \mathrm{S}(s)+2 \mathrm{Al}(s) \rightleftharpoons 6 \mathrm{Ag}(s)+3 \mathrm{S}^{2-}(a q)+2 \mathrm{Al}^{3+}(a q)$$ a. Using data in Appendix \(4,\) calculate \(\Delta G^{\circ}, K,\) and \(\mathscr{E}^{\circ}\) for the above reaction at \(25^{\circ} \mathrm{C} .\left[\text { For } \mathrm{Al}^{3+}(a q), \Delta G_{\mathrm{f}}^{\circ}=-480 . \mathrm{kJ} / \mathrm{mol.}\right]\) b. Calculate the value of the standard reduction potential for the following half-reaction: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and \(1.0 \times 10^{-4} M(\text { left side })\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+}\).)

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C} :\) $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu}$$ The mass of each electrode is 200. g. a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after 10.0 A of current has flowed for 10.0 h. (Assume each half-cell contains 1.00 L of solution.) c. Calculate the mass of each electrode after 10.0 h. d. How long can this battery deliver a current of 10.0 A before it goes dead?

A galvanic cell is based on the following half-reactions: $$\begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{aligned}$$ In this cell, the silver compartment contains a silver electrode and excess AgCl(s) \(\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) . a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) . b. Assuming 1.0 \(\mathrm{L}\) of 2.0\(M \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of 0.52 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3} )\) . $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K=1.0 \times 10^{13}$$

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