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From these data, $$ \begin{aligned} \mathrm{S}(\text { rhombic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-296.06 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{S}(\text { monoclinic })+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{SO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-296.36 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate the enthalpy change for the transformation $$ S \text { (rhombic) } \longrightarrow \mathrm{S} \text { (monoclinic) } $$ (Monoclinic and rhombic are different allotropic forms of elemental sulfur.)

Short Answer

Expert verified
The enthalpy change for the transformation of sulfur from its rhombic form to its monoclinic form is -0.30 kJ/mol

Step by step solution

01

Apply Hess's Law

According to Hess's Law, total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place. So, if we can create a sequence of reactions that add up to the target reaction, the sum of the enthalpies for those reactions will be the enthalpy change for the target reaction.
02

Write out the given reactions

\[\mathrm{S}(\text { rhombic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \ \Delta H_{\mathrm{rxn}}^{\circ} =-296.06 \mathrm{~kJ} / \mathrm{mol} \ \mathrm{S}(\text { monoclinic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \ \Delta H_{\mathrm{rxn}}^{\circ} =-296.36 \mathrm{~kJ} / \mathrm{mol}\]
03

Manipulate the given reactions to create the target reaction

The target reaction can be obtained by subtracting the reaction involving rhombic sulfur from the one involving monoclinic sulfur. Keeping in mind that the enthalpy change for a reverse reaction is of equal magnitude but opposite sign, we will reverse the first reaction before subtracting. This can be represented as follows: \[\mathrm{S}(\text { monoclinic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \ \mathrm{SO}_{2}(g) \longrightarrow \mathrm{S}(\text { rhombic })+\mathrm{O}_{2}(g)\] Resulting in: \[\mathrm{S}(\text { rhombic }) \longrightarrow \mathrm{S}(\text { monoclinic }) \]
04

Calculate the overall enthalpy change

Since the individual enthalpy changes for the two reactions are \(-296.36 kJ/mol\) and \(296.06 kJ/mol\), the enthalpy for the target reaction is their sum when following Hess's Law, that is, \(-296.36 kJ/mol + 296.06 kJ/mol = -0.30 kJ/mol\).

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Key Concepts

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

Hess's Law
Understanding Hess's Law is critical for calculating the enthalpy changes of chemical reactions that don't occur directly.

It states that the total enthalpy change during the complete course of a chemical reaction is the same whether the reaction is made in one step or several steps. It is a manifestation of the conservation of energy, specifically for thermodynamic processes.

The beauty of Hess's Law lies in its ability to simplify the calculation of enthalpy changes for complex reactions. By breaking down the reactions into individual, known steps, we can use algebra to add or subtract the known enthalpy changes to derive the unknown ones.

An easy way to utilize Hess's Law is to think of lego blocks. Imagine you want to build a certain model, and it doesn’t matter if you assemble it in one specific order, or if you put the pieces together randomly. The final model (or reaction product) and its total energy (or enthalpy) will be the same, as long as all the same pieces (or reaction steps) are used.
Allotropic Forms
Allotropes are different forms of the same element, where atoms are bonded together in different ways. For example, carbon can exist as diamond, graphite, or graphene, among other forms. Each allotrope has distinct properties, including different enthalpies.

In the exercise, rhombic and monoclinic sulfur are allotropic forms of sulfur. Rhombic sulfur is the most stable form at room temperature, while monoclinic sulfur is stable at higher temperatures. Even though they are the same element, their differing crystal structures result in different reactions with oxygen.

When calculating enthalpy changes involving different allotropic forms, it's important to consider that these forms may react differently and thus have different enthalpies. This is key in ensuring the accuracy of thermodynamic calculations, and understanding this allows students to grasp why the final enthalpy change for the transition between these two forms is not zero, despite both being sulfur.
Chemical Thermodynamics
Chemical Thermodynamics is the branch of chemistry that deals with the relationships between chemical reactions and energy changes involving heat. It's the underlying science that helps us understand spontaneous processes and the direction in which chemical reactions proceed.

In this context, enthalpy change is a quantity that describes the heat flow in reactions at constant pressure. It's an essential piece of a larger puzzle that includes entropy and free energy, determining whether a reaction is energetically favorable.

When solving problems in thermodynamics, it's important to always pay attention to units, as well as the signs of the enthalpy changes (exothermic reactions release energy and thus have a negative enthalpy change, while endothermic reactions absorb energy and have a positive enthalpy change).

Through the practical application of Hess's Law, as seen in the given exercise, chemical thermodynamics provides powerful tools for predicting feasible reactions and designing chemical processes efficiently.

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Most popular questions from this chapter

A piece of silver of mass \(362 \mathrm{~g}\) has a heat capacity of \(85.7 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). What is the specific heat of silver?

Lime is a term that includes calcium oxide \((\mathrm{CaO}\) also called quicklime) and calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2},\right.\) also called slaked lime \(] .\) It is used in the steel industry to remove acidic impurities, in airpollution control to remove acidic oxides such as \(\mathrm{SO}_{2}\), and in water treatment. Quicklime is made industrially by heating limestone \(\left(\mathrm{CaCO}_{3}\right)\) above \(2000^{\circ} \mathrm{C}\) : $$ \begin{aligned} \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \\ \Delta H^{\circ} &=177.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Slaked lime is produced by treating quicklime with water: $$ \begin{aligned} \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) \\ \Delta H^{\circ} &=-65.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime \(\left(0.946 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) and slaked lime \(\left(1.20 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a 500 -g sample of water reacts with an equimolar amount of \(\mathrm{CaO}\) (both at an initial temperature of \(\left.25^{\circ} \mathrm{C}\right)\), what is the final temperature of the product, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) Assume that the product absorbs all of the heat released in the reaction. (b) Given that the standard enthalpies of formation of \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are \(-635.6 \mathrm{~kJ} / \mathrm{mol}\) and \(-285.8 \mathrm{~kJ} / \mathrm{mol}\), respectively, cal- culate the standard enthalpy of formation of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Stoichiometry is based on the law of conservation of mass. On what law is thermochemistry based?

Portable hot packs are available for skiers and people engaged in other outdoor activities in a cold climate. The air-permeable paper packet contains a mixture of powdered iron, sodium chloride, and other components, all moistened by a little water. The exothermic reaction that produces the heat is a very common one- the rusting of iron: $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ When the outside plastic envelope is removed, \(\mathrm{O}_{2}\) molecules penetrate the paper, causing the reaction to begin. A typical packet contains \(250 \mathrm{~g}\) of iron to warm your hands or feet for up to \(4 \mathrm{~h}\). How much heat (in \(\mathrm{kJ}\) ) is produced by this reaction? (Hint: See Appendix 2 for \(\Delta H_{\mathrm{f}}^{\circ}\) values.

Consider the following data: $$ \begin{array}{lcc} \text { Metal } & \text { Al } & \text { Cu } \\ \hline \text { Mass (g) } & 10 & 30 \\ \text { Specific heat }\left(\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right) & 0.900 & 0.385 \\ \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & 40 & 60 \end{array} $$ When these two metals are placed in contact, which of the following will take place? (a) Heat will flow from Al to Cu because Al has a larger specific heat. (b) Heat will flow from Cu to Al because Cu has a larger mass. (c) Heat will flow from Cu to Al because Cu has a larger heat capacity. (d) Heat will flow from Cu to Al because Cu is at a higher temperature. (e) No heat will flow in either direction.

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