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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

Given $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=3351.4 \mathrm{~kJ} $$ (a) What is the heat of formation of aluminum oxide? (b) What is \(\Delta H^{\circ}\) for the formation of \(12.50 \mathrm{~g}\) of aluminum oxide?

Natural gas companies in the United States use the "therm" as a unit of energy. One therm is \(1 \times 10^{5} \mathrm{BTU}\). (a) How many joules are in one therm? \(\left(1 \mathrm{~J}=9.48 \times 10^{-4} \mathrm{BTU}\right)\) (b) When propane gas, \(\mathrm{C}_{3} \mathrm{H}_{8}\), is burned in oxygen, \(\mathrm{CO}_{2}\) and steam are produced. How many therms of energy are given off by \(1.00 \mathrm{~mol}\) of propane gas?

In the late eighteenth century Priestley prepared ammonia by reacting \(\mathrm{HNO}_{3}(g)\) with hydrogen gas. The thermodynamic equation for the reaction is $$ \mathrm{HNO}_{3}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-637 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) when one mole of hydrogen gas reacts. (b) What is \(\Delta H\) when \(10.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g)\) is made to react with an excess of steam to form \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{2}\) gases?

Calculate (a) \(g\) when a system does \(54 \mathrm{~J}\) of work and its energy decreases by \(72 \mathrm{~J}\). (b) \(\Delta E\) for a gas that releases \(38 \mathrm{~J}\) of heat and has \(102 \mathrm{~J}\) of work done on it.

Stainless steel accessories in cars are usually plated with chromium to give them a shiny surface and to prevent rusting. When \(5.00 \mathrm{~g}\) of chromium at \(23.00^{\circ} \mathrm{C}\) absorbs \(62.5 \mathrm{~J}\) of heat, the temperature increases to \(50.8^{\circ} \mathrm{C}\). What is the specific heat of chromium?
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