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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

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

Step by step solution

01

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.
02

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: $$ \mathrm{PbS}(s)+\frac{3}{2} \mathrm{O}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{Pb}(s)+\mathrm{SO}_{2}(g)+\mathrm{CO}(g) $$
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A typical fat in the body is glyceryl trioleate, \(\mathrm{C}_{57} \mathrm{H}_{104} \mathrm{O}_{6}\). When it is metabolized in the body, it combines with oxygen to produce carbon dioxide, water, and \(3.022 \times 10^{4} \mathrm{~kJ}\) of heat per mole of fat. (a) Write a balanced thermochemical equation for the metabolism of fat. (b) How many kilojoules of energy must be evolved in the form of heat if you want to get rid of five pounds of this fat by combustion? (c) How many nutritional calories is this? (1 nutritional calorie = \(1 \times 10^{3}\) calories)

Titanium is a metal used in jet engines. Its specific heat is \(0.523 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If \(5.88 \mathrm{~g}\) of titanium absorbs \(4.78 \mathrm{~J}\), what is the change in temperature?

Calcium carbide, \(\mathrm{CaC}_{2}\), is the raw material for the production of acetylene (used in welding torches). Calcium carbide is produced by reacting calcium oxide with carbon, producing carbon monoxide as a byproduct. When one mole of calcium carbide is formed, \(464.8 \mathrm{~kJ}\) is absorbed. (a) Write a thermochemical equation for this reaction. (b) Is the reaction exothermic or endothermic? (c) Draw an energy diagram showing the path of this reaction. (Figure \(8.4\) is an example of such an energy diagram.) (d) What is \(\Delta H\) when \(1.00 \mathrm{~g}\) of \(\mathrm{CaC}_{2}(\mathrm{~g})\) is formed? (e) How many grams of carbon are used up when \(20.00 \mathrm{~kJ}\) of heat is absorbed?

Given the following thermochemical equations $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-571.6 \mathrm{~kJ} \\ \mathrm{~N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(l) & & \Delta H=-73.7 \mathrm{~kJ} \\ \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{O}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{HNO}_{3}(l) & & \Delta H=-174.1 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), is used in the manufacture of resins and glues. When \(5.00 \mathrm{~g}\) of urea is dissolved in \(250.0 \mathrm{~mL}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\) at \(30.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, \(27.6 \mathrm{~kJ}\) of heat is absorbed. (a) Is the solution process exothermic? (b) What is \(q_{\mathrm{H}_{2} \mathrm{O}}\) ? (c) What is the final temperature of the solution? (Specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) (d) What are the initial and final temperatures in \({ }^{\circ} \mathrm{F}\) ?

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