Question

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 in the 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

To produce 150 kg of 1,4-dicyanobutane (NC-(CH2)4-CN) per hour, the required current is approximately 72,935.48 A.

Step-by-step solution

Determine the molecular weight of 1,4-dicyanobutane

To calculate the moles of 1,4-dicyanobutane produced, we first need the molecular weight. The molecular formula for 1,4-dicyanobutane is NC-(CH2)4-CN. The molecular weight of NC-(CH2)4-CN: = (12.01 * 4) + (2 * 14.01) + (1.01 * 10) + (2 * 12.01) = 48.04 + 28.02 + 10.1 + 24.02 = 110.18 g/mol

Calculate the moles of 1,4-dicyanobutane

Now we need to find out how many moles are present in 150 kg of 1,4-dicyanobutane. To do this, we'll convert kg to g and then use the molecular weight to find the moles. Mass = 150 kg = 150,000 g Moles of 1,4-dicyanobutane = mass (g) / molecular weight = 150,000 g / 110.18 g/mol = 1361.6 mol

Calculate the moles of electrons involved

The balanced reduction reaction is given as: 2 CH2=CHCN + 2 H+ + 2 e- → NC-(CH2)4-CN We can see that 2 moles of electrons are involved in the production of 1 mole of 1,4-dicyanobutane. Therefore, to calculate the moles of electrons involved, we can multiply the moles of 1,4-dicyanobutane by 2. Moles of electrons = 2 × 1361.6 mol = 2723.2 mol

Calculate the charge required

Next, we'll use Faraday's law to calculate the charge required to produce the desired amount of 1,4-dicyanobutane. Faraday's constant (F) is approximately 96,485 C/mol. Charge (Q) = moles of electrons × Faraday's constant = 2723.2 mol × 96,485 C/mol = 262,569,722 C

Calculate the current

Now we'll find the current (I) required to produce 150 kg of 1,4-dicyanobutane per hour. We'll convert the charge-per-mol value calculated above to charge-per-hour and then use the formula I = Q/t. Charge per hour = 262,569,722 C Time (t) = 1 hour = 3600 seconds Current (I) = Q/t = 262,569,722 C / 3600 s = 72,935.48 A Therefore, the required current to produce 150 kg of 1,4-dicyanobutane per hour is approximately 72,935.48 A.

Key concepts

Faraday's law

Faraday's Law of Electrolysis is fundamental in understanding electrochemical reactions. This law states that the amount of substance that undergoes oxidation or reduction at an electrode during electrolysis is directly proportional to the total electric charge passed through the system. Essentially, Faraday's law meshes chemistry with electricity.

For any given substance, the law is summarized by the formula: \( Q = n \times F \) where:

• \( Q \) is the total electric charge (in coulombs),
• \( n \) is the number of moles of electrons exchanged in the reaction,
• \( F \) is Faraday's constant, approximately 96,485 coulombs per mole of electrons.

In the context of this problem, Faraday's law helps determine how much charge is required to produce a specified amount of a chemical product. This is achieved by calculating how many moles of electrons are involved and subsequently applying the constant to find the total charge. Understanding this principle allows industries to design processes like the electrochemical production of organic compounds efficiently.

Molecular weight calculation

Calculating molecular weight is crucial for determining how much of a particular substance is produced in a reaction. The molecular weight is the sum of the atomic weights of every atom in a molecule's chemical formula.

Consider 1,4-dicyanobutane as an example. To find its molecular weight, we add up the atomic weights of all the involved atoms:

• Carbon (C), with an atomic weight of 12.01, appears six times.
• Nitrogen (N), with an atomic weight of 14.01, appears twice.
• Hydrogen (H), with an atomic weight of 1.01, appears ten times.

Plug these numbers into the formula to get: \( \text{Molecular Weight} = (12.01 \times 6) + (14.01 \times 2) + (1.01 \times 10) = 110.18 \text{ g/mol} \)
Knowing the molecular weight, you can convert mass to moles (or vice versa), which is essential in chemical quantification and process scaling. In this case, it allows for determining how much 1,4-dicyanobutane results from a reaction.

Nylon production

The production of nylon is a multi-step process in the industrial world. Nylon itself is a type of polymer known for its strength and elasticity. The process often begins with the creation of intermediary compounds like 1,4-dicyanobutane, which undergoes further chemical transformation.

For nylon production, 1,4-dicyanobutane is reduced chemically using hydrogen gas, transforming it into 1,6-hexanediamine. This compound is crucial because it acts as one of the monomers, or building blocks, which are combined in polymerization reactions to form nylon.

The use of electrochemical reactions in these processes is not only innovative but also efficient. It reduces the dependency on traditional chemical processes that might require higher energy inputs or produce more emissions, making electrochemical routes an attractive alternative in the context of sustainable industrial practices.

Reduction reactions

Reduction reactions are a core aspect of electrochemical processes. These reactions involve the gain of electrons by a chemical species. In an electrochemical cell, reduction occurs at the cathode, one of the electrodes.

The given reaction: \( 2 \text{CH}_2=\text{CHCN} + 2 \text{H}^+ + 2 \text{e}^- \rightarrow \text{NC}-(\text{CH}_2)_4-\text{CN} \) is a reduction reaction where electrons are added to the reactants, converting them into the product, 1,4-dicyanobutane.

The understanding of reduction reactions is vital for predicting the outcome of electrochemical processes. In this context, knowing that 2 moles of electrons are required to produce 1 mole of the product places emphasis on accurate electron accounting throughout the reaction process, ensuring the desired quantities are produced efficiently.