Problem 57 - Chemistry: The Central Science

Question

Write a chemical equation for the reaction that occurs when \(\mathrm{PbS}\) is roasted in air. Why might a sulfuric acid plant be located near a plant that roasts sulfide ores?

Step-by-Step Solution

Answer

The balanced chemical equation for the roasting of lead sulfide in air is: \(PbS + 2 O_2 \rightarrow PbO + SO_2\) A sulfuric acid plant might be located near a plant that roasts sulfide ores because sulfur dioxide (SO2), a significant byproduct of roasting sulfide ores, is a key ingredient in the production of sulfuric acid (H2SO4). This proximity would reduce transportation costs, environmental impacts, and ensure a consistent supply of sulfur dioxide.

Step 1: Identify the reactants and products of the reaction

Since PbS is being roasted in the air, the reactants are lead sulfide (PbS) and oxygen (O2). The main product will be lead oxide (PbO) as the sulfide ion reacts with oxygen during the roasting. Additionally, sulfur dioxide (SO2) will be generated as a byproduct.

Step 2: Write the unbalanced chemical equation

Now we can write the unbalanced chemical equation for the reaction: PbS + O2 -> PbO + SO2

Step 3: Balance the chemical equation

Next, we will balance the chemical equation: PbS + 2 O2 -> PbO + SO2 This is the balanced chemical equation for the roasting of lead sulfide in air.

Step 4: Explain the connection with a sulfuric acid plant

A significant byproduct of roasting sulfide ores, as shown in the equation above, is sulfur dioxide (SO2). Sulfur dioxide is a key ingredient in the production of sulfuric acid (H2SO4) and can be converted to sulfur trioxide (SO3) by the contact process in a sulfuric acid plant. Sulfur trioxide then reacts with water to form sulfuric acid: SO3 + H2O -> H2SO4 Having a sulfuric acid plant near a plant that roasts sulfide ores would be beneficial because the sulfur dioxide produced as a byproduct could be directly utilized in the production of sulfuric acid, reducing transportation costs and environmental impacts. Additionally, it ensures a consistent and abundant supply of sulfur dioxide required to produce sulfuric acid.

Key Concepts

Chemical Equations

Understanding chemical equations is essential for studying chemical reactions. In simple terms, they are like recipes for chemists, telling them what substances react (reactants) and what substances are produced (products). For example, when lead sulfide (\textbf{PbS}) is roasted in air, it reacts with oxygen (\textbf{O}\textsubscript{2}) to produce lead oxide (\textbf{PbO}) and sulfur dioxide (\textbf{SO}\textsubscript{2}). The equation that represents this reaction is: \[ \text{PbS} + \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \] Such equations must be balanced, meaning the number of atoms for each element must be the same on both sides of the equation. This respects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

Sulfuric Acid Production

Sulfuric acid (\textbf{H}\textsubscript{2}\textbf{SO}\textsubscript{4}) is one of the most important industrial chemicals, used in various applications such as fertilizer production, mineral processing, and chemical synthesis. The primary raw material for producing sulfuric acid is sulfur dioxide (\textbf{SO}\textsubscript{2}), which can be obtained from the roasting of sulfide ores like lead sulfide (\textbf{PbS}). The chemical reaction for producing sulfuric acid from sulfur trioxide (\textbf{SO}\textsubscript{3}) is: \[ \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \] However, the process also involves converting \textbf{SO}\textsubscript{2} to \textbf{SO}\textsubscript{3}, which is addressed in the contact process.

Balancing Chemical Equations

Balancing chemical equations is a key skill in chemistry, providing insights into the quantities of reactants needed and products formed. To balance a reaction, one must ensure that the number of atoms of each element is equal on both sides of the equation. For instance, in the roasting of \textbf{PbS}, the initial unbalanced equation is: \[ \text{PbS} + \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \] After balancing, we get: \[ \text{PbS} + 2 \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \] Notice how adding a coefficient of '2' in front of \textbf{O}\textsubscript{2} balances the oxygen atoms. Learning to balance equations helps students understand the stoichiometry of reactions, that is, the proportional relationship between reactants and products in a chemical reaction.

Contact Process

The contact process is an industrial method used for manufacturing sulfuric acid at a high scale. It involves several steps starting with the oxidation of sulfur dioxide to sulfur trioxide. This is catalyzed by vanadium(V) oxide and takes place at approximately 450°C: \[ 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 \]

Sulfur Trioxide to Sulfuric Acid Conversion

Once we have sulfur trioxide (\textbf{SO}\textsubscript{3}), it is absorbed into concentrated sulfuric acid to create oleum, which is then diluted with water to produce the desired sulfuric acid: \[ \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \] \[ \text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2 \text{H}_2\text{SO}_4 \] The contact process is energy efficient and produces sulfuric acid of very high purity, which is why it's widely used in the chemical industry.