Question

When lead(II) sulfide is heated to high temperatures in a stream of pure oxygen gas, solid lead(II) oxide forms with the release of gaseous sulfur dioxide. Write the unbalanced chemical equation for this reaction.

Short answer

The balanced chemical equation for the reaction of lead(II) sulfide with pure oxygen gas to form solid lead(II) oxide and gaseous sulfur dioxide is: \(2PbS + 3O_2 \rightarrow 2PbO + 2SO_2\).

Step-by-step solution

Identify the reactants

The reactants are the substances that are participating in the reaction. According to the problem statement, the reactants are lead(II) sulfide (solid) and pure oxygen gas.

Identify the products

The products are the substances that are formed as a result of the reaction. According to the problem statement, the products are solid lead(II) oxide and gaseous sulfur dioxide.

Write the unbalanced chemical equation

Now, let's write the unbalanced chemical equation. The reactants (lead(II) sulfide and pure oxygen gas) are on the left side, and the products (solid lead(II) oxide and gaseous sulfur dioxide) are on the right side. To represent these chemical species in the equation, use their chemical formulas: PbS (lead(II) sulfide), O2 (oxygen gas), PbO (lead(II) oxide), and SO2 (sulfur dioxide). Unbalanced chemical equation: \(PbS + O_2 \rightarrow PbO + SO_2\)

Balance the chemical equation

To balance the chemical equation, we need to equalize the number of atoms of each element on both sides of the equation. In the unbalanced equation, \(PbS + O_2 \rightarrow PbO + SO_2\) There is 1 atom of Pb, 1 atom of S, and 2 atoms of O on the left side. There is 1 atom of Pb, 1 atom of S, 1 atom of O in PbO, and 2 atoms of O in SO2 on the right side. The number of Pb and S atoms are already balanced. We need to balance the number of O atoms. The balanced equation is: \(PbS + \frac{3}{2}O_2 \rightarrow PbO + SO_2\) As we do not commonly use fractions in chemical equations, we will get whole numbers by multiplying the equation by 2: \(2PbS + 3O_2 \rightarrow 2PbO + 2SO_2\) We can now see that the chemical equation is balanced, with 2 atoms of Pb, 2 atoms of S, and 6 atoms of O on both sides of the equation.

Key concepts

Understanding Lead(II) Sulfide

Lead(II) sulfide, represented by the chemical formula \( PbS \), is an important inorganic compound. It is typically solid and appears in nature as the mineral galena, which is also the most common ore of lead. Lead(II) sulfide features prominently in educational chemistry due to its role in various chemical reactions.

• Lead atoms carry a +2 charge, while sulfide ions carry a -2 charge.
• This charge balance results in the formation of lead(II) sulfide.

This compound is a key reactant in numerous chemical processes, especially those related to metallurgy and the production of lead. In chemical reactions, its transformation often involves changes in both its physical and chemical properties. Understanding \( PbS \) sets the stage for exploring complex chemical interactions.

Oxidation Reaction Fundamentals

Oxidation reactions are a crucial aspect of chemistry, involving the transfer of electrons between substances. In the context of lead(II) sulfide, oxidation occurs when the substance reacts with oxygen. This process results in the conversion of \( PbS \) to \( PbO \) (lead(II) oxide) and \( SO_2 \) (sulfur dioxide).

• Oxidation is characterized by the gain of oxygen or the loss of electrons.
• It coincides with reduction, the loss of oxygen, or the gain of electrons by another substance.

Such reactions are essential for understanding transitions in elements and compounds. In our example, lead(II) sulfide's oxidation demonstrates how energy in the form of heat can drive chemical and structural transformations.

The Principle of Stoichiometry

Stoichiometry is the bridge that connects the qualitative aspect of chemical reactions to the quantitative one. It involves measuring and calculating the quantities of reactants and products in a chemical reaction. By understanding stoichiometry, students can predict how much product will form from a given amount of reactant, and vice versa.

• Stoichiometry is based on the law of conservation of mass.
• It ensures that the number of atoms for each element is equal on both sides of a chemical equation.

In our balanced reaction between lead(II) sulfide and oxygen, stoichiometry guides us to use the correct proportions: \(2PbS + 3O_2 \rightarrow 2PbO + 2SO_2\). These coefficients represent the molar ratios, reflecting real-world applications of chemical theory to practical problems.

Exploring Chemical Reactions

Chemical reactions are processes where reactants transform into products, involving the making and breaking of chemical bonds. They can be exothermic, releasing energy, or endothermic, absorbing energy. Our example with \( PbS \) is an exothermic reaction, primarily due to the release of sulfur dioxide gas.

• Chemical reactions must always conform to the principles of stoichiometry.
• They may occur in different stages, such as synthesis, decomposition, single and double displacement.

Recognizing the type of reaction helps predict the structure of the products and identify potential energy changes. Chemical reactions are foundational to the transformation of materials and are crucial in both natural processes and industrial applications.