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Chapter 8: Problem 33

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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Most popular questions from this chapter

Equal masses of liquid A, initially at \(100^{\circ} \mathrm{C}\), and liquid B, initially at \(50^{\circ} \mathrm{C}\), are combined in an insulated container. The final temperature of the mixture is \(80^{\circ} \mathrm{C}\). All the heat flow occurs between the two liquids. The two liquids do not react with each other. Is the specific heat of liquid \(A\) larger than, equal to, or smaller than the specific heat of liquid B?

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 \(1 \mathrm{~atm}\).

Use the appropriate table to calculate \(\Delta H^{\circ}\) for (a) the reaction between copper(II) oxide and carbon monoxide to give copper metal and carbon dioxide. (b) the decomposition of one mole of methyl alcohol (CH \(_{3} \mathrm{OH}\) ) to methane and oxygen gases.

Nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\), is an explosive most often used in mine or quarry blasting. It is a powerful explosive because four gases \(\left(\mathrm{N}_{2}\right)\) \(\mathrm{O}_{2}, \mathrm{CO}_{2}\), and steam) are formed when nitroglycerine is detonated. In addition, \(6.26 \mathrm{~kJ}\) of heat is given off per gram of nitroglycerine detonated. (a) Write a balanced thermochemical equation for the reaction. (b) What is \(\Delta H\) when \(4.65\) mol of products is formed?

Isooctane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), is a component of gasoline. When \(0.500 \mathrm{~g}\) of isooctane is burned, \(24.06 \mathrm{~kJ}\) of heat is given off. If \(10.00 \mathrm{mg}\) of isooctane is burned in a bomb calorimeter (heat capacity \(=5175 \mathrm{~J} /{ }^{\circ} \mathrm{C}\) ) initially at \(23.6^{\circ} \mathrm{C}\), what is the temperature of the calorimeter when reaction is complete?
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