Question

The following reactions do not occur in aqueous solutions. Balance their equations by the half-equation method, as suggested in Are You Wondering \(5-2\) (a) \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \longrightarrow\) \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

Short answer

The balanced equations are: (a): \(\mathrm{CH}_{4}(\mathrm{g})+2\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+2\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), (b): \(8\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+8\mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{S}_{8}(\mathrm{s})+8\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), (c): \(3\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+8\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+6\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

Step-by-step solution

Balancing Reaction (a)

For the reaction \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), start by balancing the carbon atoms, followed by the nitrogen atoms and finally the hydrogen atoms. This yields the balanced equation \(\mathrm{CH}_{4}(\mathrm{g})+2\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+2\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

Balancing Reaction (b)

For the reaction \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), we first balance the sulfur atoms, then the hydrogen atoms, and finally the oxygen atoms. The balanced equation is thus \(8\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+8\mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{S}_{8}(\mathrm{s})+8\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

Balancing Reaction (c)

For the reaction \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\), the balancing starts with nitrogen, then hydrogen, and finally chlorine and oxygen. The balanced equation becomes \(3\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+8\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+6\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\).

Key concepts

Half-Reaction Method

The Half-Reaction method is a structured approach to balancing redox reactions, where the oxidation and reduction parts of a chemical reaction are separated into two distinct half-reactions. Each half-reaction helps you visualize how electrons are transferred between different species.
Balancing each half-reaction involves ensuring that the number of electrons lost in oxidation equals the number gained in reduction.

• First, write the individual oxidation and reduction half-reactions.
• Then balance each half-reaction for atoms and charge by adding electrons, water molecules, or hydrogen ions as needed.
• Finally, add the two half-reactions back together, making sure that the overall number of electrons cancels out.

This method allows for the systematic and clear balancing of complex reactions involving electron transfers. It can be applied to many reactions, even when they don't occur in aqueous solutions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical changes involving the transfer of electrons between two species. These reactions are essential in both biological and industrial processes.

• In a redox reaction, one substance loses electrons (oxidation) while another substance gains electrons (reduction).
• The substance that donates electrons is called the reducing agent, and the one that accepts electrons is the oxidizing agent.

Understanding redox reactions is pivotal because they are responsible for generating energy in batteries, driving complex biological processes in cells, and even forming the basis of rusting in metals. The balance in these reactions is achieved by equalizing the number of electrons gained and lost, ensuring mass and charge balance.

Chemical Stoichiometry

Chemical stoichiometry involves the calculation of reactants and products in chemical reactions, ensuring that the law of conservation of mass is upheld. It plays a key role in understanding reaction quantities and relates closely to balancing equations.
In stoichiometry:

• Coefficients in a balanced equation tell us the relative amounts of reactants and products, allowing for precise predictions of mass and volume relationships.
• This is important when comparing proportions of different substances involved in reactions or when calculating yields.
• It helps achieve the correct ratio, ensuring no excess reactants or incomplete reactions.

Stoichiometry is the backbone for analyzing how molecules react with each other, providing clarity and precision to chemical processes.