Question

Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. (b) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, \(\mathrm{PbS}(\mathrm{s})\), reacts with ozone to form \(\mathrm{PbSO}_{4}(s)\) and \(\mathrm{O}_{2}(g) .\) (d) When heated in air, \(\mathrm{ZnS}(s)\) is converted to \(\mathrm{ZnO}\). (e) Potassium peroxide reacts with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\).

Short answer

The short answer is as follows: a) \(\mathrm{2HgO\rightarrow 2Hg + O_2}\) b) \(\mathrm{2Cu(NO_3)_2\rightarrow 2CuO + 4NO_2 + O_2}\) c) \(\mathrm{PbS + 2O_3\rightarrow PbSO_4 + O_2}\) d) \(\mathrm{ZnS + O_2\rightarrow ZnO + SO_2}\) e) \(\mathrm{2K_2O_2 + 2CO_2\rightarrow 2K_2CO_3 + O_2}\)

Step-by-step solution

a) Decomposition of mercury(II) oxide

In this reaction, mercury(II) oxide decomposes into oxygen gas and mercury metal upon heating. The reactant is mercury(II) oxide (\(\mathrm{HgO}\)), and the products are oxygen gas (\(\mathrm{O_2}\)) and mercury metal (\(\mathrm{Hg}\)). Now we need to balance the equation to ensure that the number of atoms of each element is the same on both sides: $$2\mathrm{HgO}\rightarrow 2\mathrm{Hg}+\mathrm{O_2}$$ The balanced equation for the decomposition of mercury(II) oxide is: $$\mathrm{2HgO}\rightarrow \mathrm{2Hg}+\mathrm{O_2}$$

b) Decomposition of copper(II) nitrate

When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide (\(\mathrm{CuO}\)), nitrogen dioxide (\(\mathrm{NO_2}\)), and oxygen gas (\(\mathrm{O_2}\)). The reactant is copper(II) nitrate (\(\mathrm{Cu(NO_3)_2}\)). Let's balance the equation: $$2\mathrm{Cu(NO_3)_2}\rightarrow 2\mathrm{CuO} + 4\mathrm{NO_2} + \mathrm{O_2}$$ The balanced equation for the decomposition of copper(II) nitrate is: $$\mathrm{2Cu(NO_3)_2}\rightarrow \mathrm{2CuO} + \mathrm{4NO_2} + \mathrm{O_2}$$

c) Reaction between lead(II) sulfide and ozone

Lead(II) sulfide, \(\mathrm{PbS(s)}\), reacts with ozone to form \(\mathrm{PbSO_4(s)}\) and \(\mathrm{O_2(g)}\). The reactants are lead(II) sulfide (\(\mathrm{PbS}\)) and ozone (\(\mathrm{O_3}\)), and the products are lead(II) sulfate (\(\mathrm{PbSO_4}\)) and oxygen gas (\(\mathrm{O_2}\)). Balancing the equation gives: $$\mathrm{PbS} + 2\mathrm{O_3}\rightarrow \mathrm{PbSO_4} + \mathrm{O_2}$$ The balanced equation for the reaction between lead(II) sulfide and ozone is: $$\mathrm{PbS} + \mathrm{2O_3}\rightarrow \mathrm{PbSO_4} + \mathrm{O_2}$$

d) Conversion of \(\mathrm{ZnS}\) in air

When heated in air, \(\mathrm{ZnS(s)}\) is converted to \(\mathrm{ZnO}\). The reactants are zinc sulfide (\(\mathrm{ZnS}\)) and oxygen gas from air (\(\mathrm{O_2}\)) and the product is zinc oxide (\(\mathrm{ZnO}\)). Balancing the equation gives: $$\mathrm{ZnS} + \mathrm{O_2}\rightarrow \mathrm{ZnO} + \mathrm{SO_2}$$ The balanced equation for the conversion of \(\mathrm{ZnS}\) in air is: $$\mathrm{ZnS} + \mathrm{O_2}\rightarrow \mathrm{ZnO} + \mathrm{SO_2}$$

e) Reaction between potassium peroxide and \(\mathrm{CO_2}\)

Potassium peroxide (\(\mathrm{K_2O_2}\)) reacts with \(\mathrm{CO_2(g)}\) to give potassium carbonate (\(\mathrm{K_2CO_3}\)) and \(\mathrm{O_2}\). Balancing the equation yields: $$2\mathrm{K_2O_2} + 2\mathrm{CO_2}\rightarrow 2\mathrm{K_2CO_3} + \mathrm{O_2}$$ The balanced equation for the reaction between potassium peroxide and \(\mathrm{CO_2}\) is: $$\mathrm{2K_2O_2} + \mathrm{2CO_2}\rightarrow \mathrm{2K_2CO_3} + \mathrm{O_2}$$

Key concepts

Decomposition Reactions

Decomposition reactions are a type of chemical reaction in which a single compound breaks down into two or more simpler substances. This process often requires an external energy source, such as heat, to overcome the chemical bonds holding the compound's atoms together.
For example, when mercury(II) oxide (\(\mathrm{HgO}\)) is heated, it decomposes into mercury metal (\(\mathrm{Hg}\)) and oxygen gas (\(\mathrm{O_2}\)). In this breakdown:

• The compound loses its structure, forming separate elements or smaller compounds.
• These reactions are essential in various chemical processes, such as metal extraction and thermal processing.

The equation for the decomposition of mercury(II) oxide illustrates the conversion of one reactant into multiple products, showcasing a classic decomposition reaction: \[2\mathrm{HgO} \rightarrow 2\mathrm{Hg} + \mathrm{O_2}\] This example is a straightforward depiction of a decomposition reaction, commonly occurring under heat.

Balancing Equations

Balancing chemical equations is vital to represent the conservation of mass, ensuring there are an equal number of atoms for each element on both sides of the equation. It's like balancing a scale, where the atoms on either side must match to reflect the reality of matter conservation in chemical reactions.
To balance an equation:

• Count the atoms for each element in both reactants and products.
• Adjust the coefficients (numbers in front of compounds) to equalize the atom counts on both sides.

For example, in the decomposition of copper(II) nitrate, balancing aligns reactants and products:\[2\mathrm{Cu(NO_3)_2} \rightarrow 2\mathrm{CuO} + 4\mathrm{NO_2} + \mathrm{O_2}\] Here, fixing the coefficients ensures all elements account equally. Copper, nitrogen, and oxygen balances result in a correct depiction of this reaction. Balancing clarifies how substances transform through chemical equations.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between substances, which changes their oxidation states. One substance loses electrons (oxidation), while another gains electrons (reduction). This shift is crucial in various chemical processes, such as metabolism and combustion.
For instance, lead(II) sulfide (\(\mathrm{PbS}\)) reacting with ozone (\(\mathrm{O_3}\)) involves lead transferring electrons:

• Lead in \(\mathrm{PbS}\) is oxidized to lead in \(\mathrm{PbSO_4}\).
• Ozone is reduced to oxygen.

The equation is:\[\mathrm{PbS} + 2\mathrm{O_3} \rightarrow \mathrm{PbSO_4} + \mathrm{O_2}\] This scenario presents a clear example of electron exchange, making redox reactions central to explaining energy changes and material transformations.

Thermal Decomposition

Thermal decomposition is a type of decomposition reaction where heat causes a compound to break into simpler substances. This process is significant in various industrial and chemical settings, where controlling heating conditions can yield different results.
Take the heating of copper(II) nitrate, which decomposes into copper(II) oxide, nitrogen dioxide, and oxygen:

• Heat acts as the driving force, breaking chemical bonds in the compound.
• The result is a transformation into multiple products.

This equation represents the process:\[2\mathrm{Cu(NO_3)_2} \rightarrow 2\mathrm{CuO} + 4\mathrm{NO_2} + \mathrm{O_2}\] Understanding thermal decomposition enables better control over reactions, such as creating new materials or eliminating unwanted substances in a compound.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. By using coefficients from a balanced equation, chemists can calculate the precise amounts of substances needed or produced.
The stoichiometry of a reaction helps predict:

• How much reactant is required to produce a desired product amount.
• The yield of products based on given quantities of reactants.

For example, in the reaction of potassium peroxide with carbon dioxide:\[2\mathrm{K_2O_2} + 2\mathrm{CO_2} \rightarrow 2\mathrm{K_2CO_3} + \mathrm{O_2}\] Knowing the coefficients lets us calculate how much of each substance participates in the reaction. Reaction stoichiometry is foundational in chemistry, critically useful in laboratories and industrial processes to maximize efficiency and outputs.