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Uranium hexafluoride, \(\mathrm{UF}_{6},\) is processed to produce fuel for nuclear reactors and nuclear weapons. UF \(_{6}\) can be produced in a two-step reaction. Solid uranium (IV) oxide, \(\mathrm{UO}_{2}\), is first made to react with hydrofluoric acid (HF) solution to form solid UF \(_{4}\) with water as a by-product. \(U F_{4}\) further reacts with fluorine gas to form UF \(_{6}\). (a) Write the balanced molecular equations for the conversion of \(U O_{2}\) into \(U F_{4}\) and the conversion of \(U F_{4}\) to \(U F_{6}\) (b) Which step is an acid-base reaction? (c) Which step is a redox reaction?

Short Answer

Expert verified
(a) The balanced equations for the given reactions are: \(UO_{2} + 4HF \rightarrow UF_{4} + 2H_{2}O\) and \(UF_{4} + 2F_{2} \rightarrow UF_{6}\) (b) The acid-base reaction is the first step: \(UO_{2} + 4HF \rightarrow UF_{4} + 2H_{2}O\) (c) The redox reaction is the second step: \(UF_{4} + 2F_{2} \rightarrow UF_{6}\)

Step by step solution

01

(Step 1: Write the balanced equation for the conversion of UO2 into UF4)

To find the balanced equation for the conversion of uranium(IV) oxide (UO2) into uranium(IV) fluoride (UF4) using hydrofluoric acid (HF), we need to combine the reactants and the products, which are water (H2O) in this case. The unbalanced equation is as follows: UO2 + HF → UF4 + H2O Next, we need to balance the equation. We find that we need to add a factor of 4 for the HF molecule: UO2 + 4HF → UF4 + 2H2O So the balanced equation is: UO2 + 4HF → UF4 + 2H2O
02

(Step 2: Write the balanced equation for the conversion of UF4 to UF6)

To find the balanced equation for the conversion of uranium(IV) fluoride (UF4) into uranium hexafluoride (UF6) using fluorine (F2), we need to combine the reactants and the product. The unbalanced equation will look like this: UF4 + F2 → UF6 Now, we need to balance the equation. We find that we need to add a factor of 2 for fluorine: UF4 + 2F2 → UF6 So the balanced equation for the second reaction is: UF4 + 2F2 → UF6
03

(Step 3: Identify which step is an acid-base reaction)

The acid-base reaction occurs when an acid donates a proton (H⁺) to a base. In our case, the first reaction is the acid-base reaction because hydrofluoric acid (HF) donates a proton to uranium(IV) oxide (UO2). This results in the production of water. UO2 + 4HF → UF4 + 2H2O
04

(Step 4: Identify which step is a redox reaction)

A redox reaction is a chemical reaction where the oxidation states of the atoms are changed. In our case, the second step is the redox reaction. This is because, in the reaction, the oxidation state of uranium remains the same (+4), but the oxidation state of fluorine increases from -1 (in F2) to -2 (in UF6), and the oxidation state of fluorine decreases from -1 (in UF4) to -2 (in UF6). UF4 + 2F2 → UF6

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Key Concepts

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

Uranium Hexafluoride
Uranium hexafluoride, or UF\(_6\), is a compound pivotal in the nuclear industry. Although it might sound complex, let's break it down. It's mainly used in processes related to nuclear fuel production and the enrichment of uranium.
The distinct aspect of UF\(_6\) is that it can exist as both solid and gas, depending on the pressure and temperature. This versatility makes it incredibly useful in separating uranium isotopes.
Here’s a quick look at how it’s produced:
  • First, uranium(IV) oxide (UO\(_2\)) reacts with hydrofluoric acid (HF) to create uranium(IV) fluoride (UF\(_4\)), with water as a byproduct.
  • Next, UF\(_4\) is reacted with fluorine gas (F\(_2\)) to form uranium hexafluoride (UF\(_6\)).
The production of UF\(_6\) is essential for providing uranium fuel in a form that's suitable for nuclear reactors. It plays a crucial role in ensuring that the uranium used is enriched to the desired levels.
Acid-Base Reaction
An acid-base reaction involves the transfer of protons (H\(^+\)) from an acid to a base. In the context of the exercise, this concept is demonstrated in the first step of producing uranium hexafluoride.
Uranium(IV) oxide (UO\(_2\)) interacts with hydrofluoric acid (HF) resulting in the formation of UF\(_4\).
This transformation is categorized as an acid-base reaction because:
  • The hydrofluoric acid (HF) donates protons to the uranium(IV) oxide.
  • This reaction results in the formation of water (H\(_2\)O) as a by-product, which is a key indicator of an acid-base reaction.
The balanced chemical equation for this process is: UO\(_2\) + 4HF \( \rightarrow \) UF\(_4\) + 2H\(_2\)O Recognizing this kind of reaction is important because it underlines the mechanism through which new compounds are formed and highlights the roles of acids and bases in chemical transformations.
Redox Reaction
Redox reactions, short for reduction-oxidation reactions, involve changes in the oxidation states of atoms through electron transfer. This type of reaction can be seen in the second step of uranium hexafluoride production.
The key players here are uranium(IV) fluoride (UF\(_4\)) and fluorine gas (F\(_2\)).
In this step:
  • The uranium remains in the same oxidation state but experiences a change in its surroundings.
  • The fluorine atoms in fluorine gas undergo a transformation where their oxidation states decrease as bonding occurs.
The balanced equation for this redox reaction is as follows: UF\(_4\) + 2F\(_2\) \( \rightarrow \) UF\(_6\) Understanding redox reactions is vital because they are fundamental to countless chemical processes, including metabolism, combustion, and corrosion. They explain how electron transfer can drastically change the properties and arrangements of atoms.

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Most popular questions from this chapter

Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{KOH}\) solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

An aqueous solution contains \(1.2 \mathrm{~m} M\) of total ions. (a) If the solution is \(\mathrm{NaCl}(a q),\) what is the concentration of chloride ion? (b) If the solution is \(\mathrm{FeCl}_{3}(a q),\) what is the concentration of chloride ion? [Section 4.5\(]\)

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathbf{b}) \mathrm{NO}_{3}^{-}\), (d) \(\mathrm{S}^{2-}\), (c) \(\mathrm{NH}_{4}^{+}\) (e) \(\mathrm{SO}_{4}^{2-} .[\) Section 4.2\(]\)

(a) Which will have the highest concentration of sodium ions: \(0.25 \mathrm{MNaCl}, 0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3},\) or \(0.075 \mathrm{MNa}_{3} \mathrm{PO}_{4} ?(\mathbf{b})\) Which will contain the greater number of moles of sodium ion: \(20.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaHCO}_{3}\) or \(15.0 \mathrm{~mL}\) of \(0.04 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S} ?\)

(a) Calculate the molarity of a solution made by dissolving 12.5 grams of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in enough water to form exactly \(750 \mathrm{~mL}\) of solution. (b) How many moles of KBr are present in \(150 \mathrm{~mL}\) of a \(0.112 \mathrm{M}\) solution? (c) How many milliliters of \(6.1 \mathrm{MHCl}\) colution a

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