Question

A lead ore, galena, consisting mainly of lead(II) sulfide, is the principal source of lead. To obtain the lead, the ore is first heated in the air to form lead oxide. $$ \mathrm{PbS}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{PbO}(s)+\mathrm{SO}_{2}(g) \quad \Delta H=-415.4 \mathrm{~kJ} $$ The oxide is then reduced to metal with carbon. $$ \mathrm{PbO}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Pb}(s)+\mathrm{CO}(g) \quad \Delta H=+108.5 \mathrm{k}] $$ Calculate \(\Delta H\) for the reaction of one mole of lead(II) sulfide with oxygen and carbon, forming lead, sulfur dioxide, and carbon monoxide.

Short answer

Answer: The enthalpy change for the overall reaction is -306.9 kJ.

Step-by-step solution

Write down the given reactions and their enthalpy changes

We are given the following two reactions: Reaction 1: $$ \mathrm{PbS}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{PbO}(s)+\mathrm{SO}_{2}(g) \quad \Delta H_{1}=-415.4 \mathrm{~kJ} $$ Reaction 2: $$ \mathrm{PbO}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Pb}(s)+\mathrm{CO}(g) \quad \Delta H_{2}=+108.5 \mathrm{~kJ} $$ We need to combine these reactions to obtain the overall reaction.

Add the reactions according to Hess's law

According to Hess's law, if a reaction can be expressed as the sum of two or more other reactions, the enthalpy change for the overall reaction is the sum of the enthalpy changes for the individual reactions. Adding Reaction 1 and Reaction 2, we get the overall reaction: $$ \mathrm{PbS}(s)+\frac{3}{2} \mathrm{O}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{Pb}(s)+\mathrm{SO}_{2}(g)+\mathrm{CO}(g) $$

Calculate the enthalpy change for the overall reaction

To find the enthalpy change, \(\Delta H\), for the overall reaction, we need to add the enthalpy changes of the individual reactions (Hess's law): $$ \Delta H = \Delta H_{1} + \Delta H_{2} $$ Substitute the values given in the exercise into the equation: $$ \Delta H = (-415.4 \mathrm{~kJ}) + (+108.5 \mathrm{~kJ}) $$ Calculate the value of \(\Delta H\): $$ \Delta H = -306.9 \mathrm{~kJ} $$ So, the enthalpy change for the given overall reaction is -306.9 kJ.

Key concepts

Enthalpy Change

Enthalpy change, often represented by \( \Delta H \), refers to the heat content change that occurs in a chemical reaction. It's crucial to understand this concept because it helps us determine whether a reaction is releasing or absorbing energy.
An exothermic reaction releases heat, leading to a negative \( \Delta H \), as seen in the first reaction where lead(II) sulfide reacts with oxygen to form lead oxide and sulfur dioxide, \( \Delta H = -415.4 \mathrm{~kJ} \).
This means the reaction releases that amount of energy. On the other hand, an endothermic reaction absorbs heat, resulting in a positive \( \Delta H \). The reduction of lead oxide by carbon to form metallic lead and carbon monoxide has \( \Delta H = +108.5 \mathrm{~kJ} \).
Understanding how to calculate overall enthalpy using Hess's Law is fundamental. Hess's Law states the total enthalpy change in a chemical reaction is the same regardless of the number of steps in the reaction. This principle allows us to add independent reactions' enthalpy changes to find the total energy change for complex reactions.

Lead(II) Sulfide

Lead(II) sulfide, represented as \( \mathrm{PbS} \), is a key component in galena, the primary ore for extracting lead. This compound consists of lead and sulfur bonded together.
Lead(II) sulfide is a typical starting material in the process of obtaining elemental lead. It exists as a dense, dark grey mineral exhibiting metallic luster. Upon heating in the air, \( \mathrm{PbS} \) undergoes a transformation in which it reacts with oxygen to produce lead oxide \( \mathrm{PbO} \) and sulfur dioxide \( \mathrm{SO}_2 \).
This transformation is a critical step in metal extraction as it prepares the ore for further processing, mainly reduction, where \( \mathrm{PbO} \) is subsequently converted into pure lead metal. Understanding the nature of \( \mathrm{PbS} \) and its role in metallurgy is essential for students delving into chemistry, particularly those interested in industrial applications of chemical reactions.

Chemical Reactions

Chemical reactions are processes where substances (reactants) are transformed into new substances (products). They play a pivotal role in both natural processes and industrial applications.
In the context of lead extraction, two primary chemical reactions occur:

• First, lead(II) sulfide reacts with oxygen producing lead oxide and sulfur dioxide \( (\mathrm{PbS} + \frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{PbO} + \mathrm{SO}_2) \).
• Next, lead oxide reacts with carbon to yield lead metal and carbon monoxide \( (\mathrm{PbO} + \mathrm{C} \rightarrow \mathrm{Pb} + \mathrm{CO}) \).

These reactions are fundamentally characterized by reactions involving oxygen (oxidation) and reactions involving carbon (reduction). Understanding these reactions' context and outcomes is vital, particularly when dealing with materials and energy changes in a chemical setting.
Recognizing the step-by-step transformations in such processes through exercises helps learners grasp the sequence of reactions and their real-world relevance better.