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Chapter 22: Problem 25

Each of the following compounds decomposes to produce \(\mathrm{O}_{2}(\mathrm{g})\) when heated: \((\mathrm{a}) \mathrm{HgO}(\mathrm{s}) ;\) (b) \(\mathrm{KClO}_{4}(\mathrm{s})\) Write plausible equations for these reactions.

Short Answer

Expert verified
The balanced equations for the decomposition of HgO and KClO4 are: 2 HgO(s) 鈫� 2 Hg(l) + O2(g) and KClO4(s) 鈫� KCl(s) + 2 O2(g), respectively.

Step by step solution

01

Write Unbalanced Decomposition Reaction for HgO

In the decomposition of mercury(II) oxide (HgO), mercury and oxygen gas are formed. The reaction can be represented as: HgO(s) 鈫� Hg(l) + O2(g). But this is not balanced yet.
02

Balancing the Decomposition Reaction for HgO

The rule for balancing chemical equations is that the number of atoms for each element should be the same on both sides of the equation. There are 2 oxygen atoms on the right side while there is only 1 on the left side in our equation. To balance this, place a 2 in front of HgO, and 1 in front of Hg, which gives us the balanced equation is: 2 HgO(s) 鈫� 2 Hg(l) + O2(g)
03

Write Unbalanced Decomposition Reaction for KClO4

In the decomposition of potassium perchlorate (KClO4), potassium chloride and oxygen gas are formed. The reaction can be represented as: KClO4(s) 鈫� KCl(s) + O2(g). But this equation is not balanced yet.
04

Balancing the Decomposition Reaction for KClO4

Again, apply the rule that the number of atoms for each element in a chemical equation must be the same on both sides. In our equation, there are 4 oxygen atoms on the left side, but only 2 on the right side. To balance this, put a 2 in front of O2. This gives us our final balanced equation: KClO4(s) 鈫� KCl(s) + 2 O2(g).

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Key Concepts

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

Decomposition Reactions
Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. These reactions usually require an input of energy, often in the form of heat, to begin. For example, when
  • Mercury(II) oxide (HgO)
  • Potassium perchlorate (KClO鈧�)
are heated, they decompose and release oxygen gas.
During decomposition:
  • HgO splits into mercury (Hg) and oxygen gas (O鈧�).
  • KClO鈧� breaks down into potassium chloride (KCl) and oxygen gas (O鈧�).
These types of reactions demonstrate how compounds can lose their original structure to form completely different products. The necessity to add heat in these cases highlights that decomposition reactions are endothermic.
Balancing Chemical Equations
Balancing chemical equations is essential because it ensures the law of conservation of mass is maintained. This law states that mass cannot be created or destroyed. In chemical equations,
we must have the same number of each type of atom on both sides of the reaction. This process:
  • Makes chemical equations more accurate.
  • Helps us understand the proportions in which reagents combine and products form.
Take the decomposition of HgO: In the unbalanced equation, there was one oxygen atom on the left, and two on the right.
Balancing involves placing coefficients in front of compounds to equalize atom counts. For HgO, putting a 2 in front of HgO and Hg balances the equation to 2 HgO(s) 鈫� 2 Hg(l) + O鈧�(g). This exact approach can be applied to balance any equation, including that of KClO鈧�.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons between substances. While decomposition reactions like the ones for HgO and KClO鈧� don't explicitly show electron transfer,
they can involve changes in oxidation states. In redox reactions:
  • Oxidation is the loss of electrons.
  • Reduction is the gain of electrons.
  • The substance that loses electrons is oxidized, and the substance that gains electrons is reduced.
In some decomposition reactions, like in advanced chemistry contexts, breaking down into elements or simpler compounds can also involve shifts in oxidation states. Being aware of redox processes is crucial for understanding a wide variety of chemical reactions, though in our current examples, they primarily showcase simple decomposition.

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Most popular questions from this chapter

Despite the fact that it has the higher molecular mass, XeO \(_{4}\) exists as a gas at \(298 \mathrm{K},\) whereas \(\mathrm{XeO}_{3}\) is a solid. Give a plausible explanation for this observation.

Show how you would use elemental sulfur, chlorine gas, metallic sodium, water, and air to produce aqueous solutions containing (a) \(\mathrm{Na}_{2} \mathrm{SO}_{3} ;\) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} .\) [Hint: You will have to use information from other chapters as well as this one.].

The solubility of \(\mathrm{Cl}_{2}(\mathrm{g})\) in water is \(6.4 \mathrm{g} \mathrm{L}^{-1}\) at \(25^{\circ} \mathrm{C}\) Some of this chlorine is present as \(\mathrm{Cl}_{2},\) and some is found as HOCl or Cl^- For the hydrolysis reaction $$\begin{array}{c}\mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \\ \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \\\K_{c}=4.4 \times 10^{-4}\end{array}$$ For a saturated solution of \(\mathrm{Cl}_{2}\) in water, calculate \(\left[\mathrm{Cl}_{2}\right],[\mathrm{HOCl}],\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\) and \(\left[\mathrm{Cl}^{-}\right]\).

The text mentions that ammonium perchlorate is an explosion hazard. Assuming that \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) is the sole reactant in the explosion, write a plausible equation(s) to represent the reaction that occurs.

Make a general prediction about which of the halogen elements, \(\mathrm{F}_{2}, \mathrm{Cl}_{2}, \mathrm{Br}_{2},\) or \(\mathrm{I}_{2},\) displaces other halogens from a solution of halide ions. Which of the halogens is able to displace \(\mathrm{O}_{2}(\mathrm{g})\) from water? Which is able to displace \(\mathrm{H}_{2}(\mathrm{g})\) from water?
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