Question

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

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

Step-by-step solution

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) → Hg(l) + O2(g). But this is not balanced yet.

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) → 2 Hg(l) + O2(g)

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) → KCl(s) + O2(g). But this equation is not balanced yet.

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) → KCl(s) + 2 O2(g).

Key concepts

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 (KClO₄)

are heated, they decompose and release oxygen gas.
During decomposition:

• HgO splits into mercury (Hg) and oxygen gas (O₂).
• KClO₄ breaks down into potassium chloride (KCl) and oxygen gas (O₂).

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) → 2 Hg(l) + O₂(g). This exact approach can be applied to balance any equation, including that of KClO₄.

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 KClO₄ 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.