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Chapter 18: Problem 129

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

Expert verified
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

01

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
02

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
03

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
04

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
05

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.

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Most popular questions from this chapter

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{M} \mathrm{Ag}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\).a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?\)

A silver concentration cell is set up at \(25^{\circ} \mathrm{C}\) as shown below:The \(\mathrm{AgCl}(s)\) is in excess in the left compartment. a. Label the anode and cathode, and describe the direction of the electron flow. b. Determine the value of \(K_{\text {sp }}\) for \(\mathrm{AgCl}\) at \(25^{\circ} \mathrm{C}\).

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 \(200 . \mathrm{g}\). a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after \(10.0 \mathrm{~A}\) of current has flowed for \(10.0 \mathrm{~h}\). (Assume each half-cell contains \(1.00 \mathrm{~L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{~h}\). d. How long can this battery deliver a current of \(10.0 \mathrm{~A}\) before it goes dead?

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.957 \mathrm{~V}\) \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.775 \mathrm{~V}\) a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

The following standard reduction potentials have been det mined for the aqueous chemistry of indium: $$ \begin{array}{cl} \mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V} \end{array} $$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q) ?\)
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