Calculate the standard enthalpy change for the reaction $$2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)$$ given that$$\begin{array}{l}2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-822.2 \mathrm{~kJ} / \mathrm{mol}\end{array}$$

Short Answer

Expert verified
The standard enthalpy change for the reaction is \(-847.6 ~kJ/mol\).

Step by step solution

01

Identifying the required reaction

The required reaction is: \(2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)\). We should aim to achieve this reaction via a combination and rearrangement of the given reactions.
02

Adjust the given reactions to fit the desired reaction

Given two reactions:1) \(2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g)\longrightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s) \)with \(\Delta H_{\mathrm{rxn}}^{\circ}=-1669.8 \mathrm{~kJ} / \mathrm{mol} \)2) \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s) \)with \(\Delta H_{\mathrm{rxn}}^{\circ}=-822.2 \mathrm{~kJ} /\mathrm{mol}\)We can see that in the first one, aluminum is reacting with oxygen to form aluminum oxide, while we need it react with iron(III) oxide, which hints that the first reaction should be left as is. But in the second reaction, iron is reacting with oxygen to form iron(III) oxide, while it needs to be produced in the final reaction, implying it needs to be reversed.
03

Finding the enthalpies of the adjusted reactions

When we reverse a reaction, the sign of the enthalpy change also reverses. Therefore, the enthalpy for the reversed reaction would be \(+822.2~ kJ/mol\). Now we have two reactions:1) \(2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g)\longrightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s) \)with \(\Delta H_{\mathrm{rxn}}^{\circ}=-1669.8 \mathrm{~kJ} / \mathrm{mol} \)And2) \( \mathrm{Fe}_{2}\mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g)\)with \(\Delta H_{\mathrm{rxn}}^{\circ}=+822.2 \mathrm{~kJ} /\mathrm{mol}\)
04

Add adjusted reactions to get the desired reaction

Both reactions are added. The \(O_2\) on the products' side in the second reaction and the \(O_2\) on the reactants' side in the first reaction will cancel out. The resulting reaction would be: \(2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)\)
05

Calculate the enthalpy change of the desired reaction

By Hess's law, the enthalpy of the reaction will be the sum of the enthalpies of these two reactions, i.e., \(\Delta H = -1669.8~kJ + 822.2~kJ = -847.6 ~ kJ/mol\)

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Key Concepts

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

Hess's Law
Hess's Law is a fundamental principle in chemical thermodynamics that enables us to determine the enthalpy change \( \Delta H \) for a chemical reaction. It is based on the premise that the total enthalpy change for a reaction is the same, regardless of the number of steps or the path taken, as long as the initial and final conditions are the same. This is because enthalpy is a state function, which means it depends only on the initial and final states and not on the path taken between them.

For instance, if you're trying to calculate the enthalpy change of a reaction that's not been measured directly, you can use other reactions that add up to your target reaction, each with known enthalpy changes. By carefully selecting and adding together these reactions, keeping track of how the enthalpy changes with each step (such as reversing the direction or scaling the reaction), you can apply Hess's Law to arrive at the enthalpy change for the overall process. This approach is valuable for studying reactions that are too slow, too fast, or perhaps too dangerous to measure directly.
Enthalpy Change of Reactions
The enthalpy change \( \Delta H \) of a reaction represents the heat absorbed or released during a chemical reaction at constant pressure. It is an intrinsic property of chemical processes and provides us with insights into the energetics of reactions. In thermodynamics, \( \Delta H \) is considered positive for endothermic reactions, where heat is absorbed from the surroundings, and negative for exothermic reactions, where heat is released.

Calculating the enthalpy change requires understanding the stoichiometry of the reaction and the standard enthalpies of formation of reactants and products. Moreover, when reactions are manipulated, such as reversing the direction, the sign of \( \Delta H \) also reverses, and if the coefficients are scaled, the enthalpy change is scaled proportionally. This property is useful when applying Hess's Law to find the enthalpy change of a target reaction by manipulating the equations of known reactions. Importantly, tabulated standard enthalpy changes are a valuable resource for calculations, as seen in the textbook exercise solution.
Chemical Thermodynamics
Chemical thermodynamics deals with the study of energy and work related to chemical reactions and changes of states of matter. It encompasses concepts such as enthalpy, entropy, and Gibbs free energy, which all pertain to the spontaneity and feasibility of a chemical process. In thermodynamics, the total energy of a system is conserved, but it can transform from one form to another, like the transformation of potential chemical energy stored in molecular bonds into thermal energy (heat).

The enthalpy change discussed in solving the textbook exercise is a key component of this broader field. Thermodynamics allows us to predict whether a reaction will occur spontaneously based on changes in enthalpy, entropy, and temperature. By understanding these principles, one can engineer chemical processes to be more efficient, control reaction conditions, and even manipulate the direction of chemical reactions to our advantage.

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Most popular questions from this chapter

How are the standard enthalpies of an element and a compound determined?

The combustion of \(0.4196 \mathrm{~g}\) of a hydrocarbon releases \(17.55 \mathrm{~kJ}\) of heat. The masses of the products are \(\mathrm{CO}_{2}=1.419 \mathrm{~g}\) and \(\mathrm{H}_{2} \mathrm{O}=0.290 \mathrm{~g}\). (a) What is the empirical formula of the compound? (b) If the approximate molar mass of the compound is \(76 \mathrm{~g}\), calculate its standard enthalpy of formation.

The \(\Delta H_{\mathrm{f}}^{\circ}\) values of the two allotropes of oxygen, \(\mathrm{O}_{2}\) and \(\mathrm{O}_{3}\), are 0 and \(142.2 \mathrm{~kJ} / \mathrm{mol}\), respectively, at \(25^{\circ} \mathrm{C}\). Which is the more stable form at this temperature?

The internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of this process. A sample of an ideal gas is allowed to expand at constant temperature against atmospheric pressure. (a) Does the gas do work on its surroundings? (b) Is there heat exchange between the system and the surroundings? If so, in which direction? (c) What is \(\Delta U\) for the gas for this process?

Consider the dissolution of an ionic compound such as potassium fluoride in water. Break the process into the following steps: separation of the cations and anions in the vapor phase and the hydration of the ions in the aqueous medium. Discuss the energy changes associated with each step. How does the heat of solution of \(\mathrm{KF}\) depend on the relative magnitudes of these two quantities? On what law is the relationship based?

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