Problem 39 - Chemistry The Central Science

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, \(\mathrm{PbS}(s)\), 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 (f) Oxygen with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\). is converted to ozone in the upper atmosphere.

Step-by-Step Solution

Answer

(a) \(2\mathrm{HgO} \rightarrow \mathrm{O}_{2} + 2\mathrm{Hg}\) (b) \(\mathrm{Cu(NO_3)_2} \rightarrow \mathrm{CuO} + 2\mathrm{NO_2} + \frac{1}{2}\mathrm{O_2}\) (c) \(\mathrm{PbS} + 2\mathrm{O_3} \rightarrow \mathrm{PbSO_4} + 2\mathrm{O_2}\) (d) \(\mathrm{ZnS} + \frac{1}{2}\mathrm{O_2} \rightarrow \mathrm{ZnO}\) (e) \(\mathrm{K_2O_2} + \mathrm{CO_2} \rightarrow \mathrm{K_2CO_3} + \mathrm{O_2}\) (f) \(\frac{3}{2}\mathrm{O_2} \rightarrow \mathrm{O_3}\)

Step 1: a) Decomposition of mercury(II) oxide

To write the balanced equation for the decomposition of mercury(II) oxide, first write the formula for the reactant (mercury(II) oxide) and the products (oxygen and mercury metal) and then find the coefficients that balance the atoms. The formula for mercury(II) oxide is \(\mathrm{HgO}\), and mercury metal is represented by \(\mathrm{Hg}\). The decomposition reaction can be written as: $$\mathrm{HgO} \rightarrow \mathrm{O}_{2} + \mathrm{Hg}$$ To balance the equation, we need 2 \(\mathrm{HgO}\) molecules to produce 2 \(\mathrm{Hg}\) atoms and 1 \(\mathrm{O}_{2}\) molecule: $$2\mathrm{HgO} \rightarrow \mathrm{O}_{2} + 2\mathrm{Hg}$$

Step 2: (b) Decomposition of copper(II) nitrate

Copper(II) nitrate has the formula \(\mathrm{Cu(NO_3)_2}\) and decomposes into copper(II) oxide (\(\mathrm{CuO}\)), nitrogen dioxide (\(\mathrm{NO_2}\)), and oxygen (\(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{Cu(NO_3)_2} \rightarrow \mathrm{CuO} + \mathrm{NO_2} + \mathrm{O_2}$$ To balance the equation, we need 2 \(\mathrm{NO_2}\) molecules and 1 \(\mathrm{O_2}\) molecule for each \(\mathrm{Cu(NO_3)_2}\) molecule decomposed: $$\mathrm{Cu(NO_3)_2} \rightarrow \mathrm{CuO} + 2\mathrm{NO_2} + \frac{1}{2}\mathrm{O_2}$$

Step 3: (c) Reaction of lead(II) sulfide with ozone

The formula for lead(II) sulfide is \(\mathrm{PbS}\) and reacts with ozone (\(\mathrm{O_3}\)) to form lead(II) sulfate (\(\mathrm{PbSO_4}\)) and oxygen (\(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{PbS} + \mathrm{O_3} \rightarrow \mathrm{PbSO_4} + \mathrm{O_2}$$ To balance the equation, we need 2 \(\mathrm{O_3}\) molecules to produce 2 \(\mathrm{O_2}\) molecules: $$\mathrm{PbS} + 2\mathrm{O_3} \rightarrow \mathrm{PbSO_4} + 2\mathrm{O_2}$$

Step 4: (d) Conversion of zinc sulfide to zinc oxide in air

The formula for zinc sulfide is \(\mathrm{ZnS}\) and is converted to zinc oxide (\(\mathrm{ZnO}\)) when heated in air (in the presence of oxygen, \(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{ZnS} + \mathrm{O_2} \rightarrow \mathrm{ZnO}$$ To balance the equation, we need 1 \(\mathrm{O_2}\) molecule to react with 1 \(\mathrm{ZnS}\): $$\mathrm{ZnS} + \frac{1}{2}\mathrm{O_2} \rightarrow \mathrm{ZnO}$$

Step 5: (e) Reaction of potassium peroxide with carbon dioxide

The formula for potassium peroxide is \(\mathrm{K_2O_2}\) and reacts with carbon dioxide (\(\mathrm{CO_2}\)) to form potassium carbonate (\(\mathrm{K_2CO_3}\)) and oxygen (\(\mathrm{O_2}\)). The reaction can be written as: $$\mathrm{K_2O_2} + \mathrm{CO_2} \rightarrow \mathrm{K_2CO_3} + \mathrm{O_2}$$ Since all elements are already balanced, no additional coefficients are needed.

Step 6: (f) Conversion of oxygen to ozone in the upper atmosphere

The conversion of oxygen (\(\mathrm{O_2}\)) to ozone (\(\mathrm{O_3}\)) in the upper atmosphere can be represented by the equation: $$\mathrm{O_2} \rightarrow \mathrm{O_3}$$ To balance the equation, note that 3 \(\mathrm{O_2}\) molecules are needed to form 2 \(\mathrm{O_3}\) molecules: $$\frac{3}{2}\mathrm{O_2} \rightarrow \mathrm{O_3} $$

Key Concepts

Composition Reactions

Composition reactions, also known as synthesis reactions, involve combining two or more substances to form a single new compound. A classic example of a composition reaction is when elements combine to create a compound, such as magnesium (Mg) reacting with oxygen (O₂) to form magnesium oxide (MgO).

These reactions can vary greatly but generally follow this pattern:

• Element or compound + Element or compound → New compound

Understanding composition reactions is essential as they are foundational to creating complex molecules from simpler ones. To analyze these reactions, pay attention to the reactants and products, ensuring they are correctly balanced to adhere to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction.

Decomposition Reactions

Decomposition reactions are the reverse of composition reactions. In decomposition reactions, a single compound breaks down into two or more simpler substances. They typically require an input of energy, such as heat, light, or electricity, to break chemical bonds.

A basic decomposition reaction follows this general form:

• AB → A + B

For instance, the decomposition of mercury(II) oxide (HgO) when heated to form mercury (Hg) and oxygen (O₂) is a classic example. These reactions are critical in processes like metallic karst, hydrogen sources, and even our global nitrogen cycle. When balancing these reactions, ensure the number of atoms of each element is the same on both sides of the equation.

Reaction Equations

Reaction equations are a concise way of representing a chemical reaction using chemical symbols and formulas. A balanced reaction equation expresses the identities and quantities of substances involved in the reaction. The principle of the reaction equations is built on the law of conservation of mass, meaning the mass of reactants must equal the mass of products.

A balanced chemical equation is critical because it reflects exactly how molecules combine and break apart. An unbalanced equation, like \(\mathrm{H_2 + O_2 → H_2O}\), requires careful balancing to yield \(2\mathrm{H_2 + O_2 → 2H_2O}\).
Balancing equations requires adjustments of coefficients—numbers placed before compounds—to ensure the same number of atoms for each element is on both sides of the equation.

Stoichiometry

Stoichiometry is a section of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It uses the balanced chemical equation to ensure the right proportions of substances for the reaction to occur as expected. This calculation helps in understanding the scaling of reactions in both laboratory settings and industrial applications.

Stoichiometry involves

• Mole ratios derived from the coefficients in a balanced equation.
• Converting masses to moles using molar masses.
• Predicting the outcomes of reactions, like the amount of product formed or reactant consumed.

Being proficient in stoichiometry allows chemists to plan reactions efficiently, determine the reactant proportions needed, and assess the efficiency of reactions. This understanding is crucial for practical applications, such as pharmaceuticals, agriculture, and materials science.