Question

Complete and balance the equations for the following reactions. a. \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow\) b. \(\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow\)

Short answer

The balanced equations for the given reactions are as follows: a. \(3\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s)\) b. \(2\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s)\)

Step-by-step solution

a. Write the complete chemical equation.

Lithium (Li) is a metal from Group 1 (Alkali metals), and Nitrogen (N) is a nonmetal from Group 15. The metal and nonmetal will form an ionic compound. Lithium has a +1 charge and Nitrogen gas exists as N2, which means it will form Lithium Nitride, Li3N. This results in the following chemical equation: \[\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s)\] Now, we need to balance this equation.

a. Balance the chemical equation.

First, count the number of atoms for each element on both sides of the equation: On the left side, there are 1 Li and 2 N atoms. On the right side, there are 3 Li and 1 N atom. To balance the Li atoms, we place a coefficient of 3 in front of Li on the left side: \[3\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s)\] Now, we have 3 Li and 2 N atoms on both sides. The equation is balanced.

b. Write the complete chemical equation.

Rubidium (Rb) is a metal from Group 1 (Alkali metals), while sulfur (S) is a non-metal. Rb has a +1 charge and sulfur (S) has a -2 charge because it is from Group 16 (Chalcogens). They will form the ionic compound Rb2S, which is Rubidium Sulfide. The full equation is: \[\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s)\] Now, we need to balance this equation.

b. Balance the chemical equation.

First, we count the number of atoms for each element on both sides: On the left side, there is 1 Rb and 1 S atom. On the right side, there are 2 Rb atoms and 1 S atom. To balance the Rb atoms, we place a coefficient of 2 in front of Rb on the left side: \[2\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s)\] Now, we have 2 Rb and 1 S atom on both sides. The equation is balanced.

Key concepts

Lithium Nitride Formation

Lithium nitride is formed when lithium metal reacts with nitrogen gas. This is a classic example of an ionic compound formation. Lithium, a Group 1 alkali metal, tends to lose one electron to form a cation, Li\(^+\), due to its low ionization energy.
On the other hand, nitrogen, a Group 15 element, gains electrons to fulfill its octet, making it capable of forming nitride anions, N\(^{3-}\).In the reaction, lithium atoms and nitrogen molecules bond to form lithium nitride, represented by the chemical formula Li\(_3\)N.

• **Unbalanced Formula:** - Li(s) + N\(_2\)(g) → Li\(_3\)N(s)
• **Balanced Formula:** - 3Li(s) + N\(_2\)(g) → Li\(_3\)N(s)

The balanced equation ensures the number of atoms for each element is equal on both sides, adhering to the law of conservation of mass. By balancing, 3 lithium atoms pair with nitrogen to form one formula unit of lithium nitride, maintaining the stoichiometry of the reaction.

Rubidium Sulfide Formation

Rubidium sulfide is an ionic compound produced when rubidium reacts with sulfur. Being another alkali metal found in Group 1 of the periodic table, rubidium tends to form Rb\(^+\) ions by losing a single electron.
Sulfur, a Group 16 chalcogen, accepts electrons to form a sulfide anion (S\(^{2-}\)).
In this reaction, the rubidium atoms and sulfur atoms combine to form rubidium sulfide with the formula Rb\(_2\)S. To achieve electrical neutrality and a stable ionic compound, two rubidium ions are necessary to balance the charge of one sulfide ion.

• **Unbalanced Formula:** - Rb(s) + S(s) → Rb\(_2\)S(s)
• **Balanced Formula:** - 2Rb(s) + S(s) → Rb\(_2\)S(s)

Balancing ensures that two rubidium atoms react with one sulfur atom, thereby satisfying the stoichiometric requirements of the reaction and leading to a stable compound.

Ionic Compound Formation

Ionic compounds form when metals transfer electrons to non-metals, resulting in positive and negative ions that attract each other due to electrostatic forces. This type of reaction is typical for elements like lithium and rubidium, both alkali metals; and non-metals such as nitrogen and sulfur.
In the case of lithium nitride and rubidium sulfide:

• Metals (lithium, rubidium) lose electrons to form positively charged cations (Li\(^+\), Rb\(^+\)).
• Non-metals (nitrogen, sulfur) gain electrons to form negatively charged anions (N\(^{3-}\), S\(^{2-}\)).
• The resulting compound adheres to the principle of charge balance, where the total positive charge equals the total negative charge.

The process is driven by the metals' tendency to achieve a stable configuration through electron donation. Non-metals accept these electrons, stabilizing themselves by achieving a full outer electron shell.
This electron transfer results in a stable ionic lattice, characterized by its high melting point and strong bond strength due to the significant electrostatic forces between the ions.