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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 $$\begin{aligned} 2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=&-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-822.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$

Short Answer

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

Step by step solution

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01

Identify the Target Reaction

The target reaction is \(2\ \mathrm{Al}(s)+\mathrm{Fe}_{2}\mathrm{O}_{3}(s) \rightarrow 2\mathrm{Fe}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)\). Our goal is to calculate the standard enthalpy change for this reaction.
02

Examine Given Reactions

The given reactions are: \(2\ \mathrm{Al}(s) + \frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s)\) with \(\Delta H^{\circ} = -1669.8 \ \mathrm{kJ/mol}\), and \(2\ \mathrm{Fe}(s) + \frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)\) with \(\Delta H^{\circ} = -822.2 \ \mathrm{kJ/mol}\).
03

Reverse Necessary Reaction

Reverse the second given reaction to align with the target reaction: \(\mathrm{Fe}_{2}\mathrm{O}_{3}(s) \rightarrow 2\ \mathrm{Fe}(s) + \frac{3}{2}\ \mathrm{O}_{2}(g)\). This changes its enthalpy to \(+822.2 \ \mathrm{kJ/mol}\).
04

Add Reactions to Find Target Reaction

Add the modified given reactions: \[(2\ \mathrm{Al}(s) + \frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s))\] and \[(\mathrm{Fe}_{2}\mathrm{O}_{3}(s) \rightarrow 2\ \mathrm{Fe}(s) + \frac{3}{2}\ \mathrm{O}_{2}(g))\].
05

Calculate Standard Enthalpy Change

Combine the enthalpy changes of the reactions: \((-1669.8 \ \mathrm{kJ/mol}) + (822.2 \ \mathrm{kJ/mol}) = -847.6 \ \mathrm{kJ/mol}\). This is the standard enthalpy change for the target reaction.

Key Concepts

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

Standard Enthalpy of Reaction
The standard enthalpy of reaction, also denoted as \( \Delta H_{\text{rxn}}^{\circ} \), is a key concept in thermochemistry. It refers to the enthalpy change that occurs when a reaction happens under standard conditions. Standard conditions typically mean a pressure of 1 atm and a temperature of 25°C (298 K). The enthalpy change is measured in kilojoules per mole (\( \text{kJ/mol} \)).

For the reaction of aluminum with iron(III) oxide discussed in the exercise, the standard enthalpy of reaction is calculated based on other known reactions. By using enthalpy values from previously determined reactions, we can piece together the enthalpy change for a new reaction, as was done using Hess's Law. The negative value of the enthalpy change \((-847.6\ \text{kJ/mol})\) indicates that the reaction is exothermic, releasing heat as it progresses.
Hess's Law
Hess's Law is a fundamental principle in chemistry that states that the total enthalpy change of a reaction is the same, no matter whether the reaction takes place in one step or a series of steps. This law is a direct result of the fact that enthalpy is a state function, meaning it only depends on the initial and final states and not on how the process is carried out.

In the given exercise, Hess's Law is used to determine the enthalpy change for the target reaction. To do this, we leverage two known reactions and their enthalpy changes. By reversing one of the reactions and combining the two, the overall equation reflects the desired reaction. Hess's Law allows us to simply add the enthalpy changes of these steps to find the total enthalpy change for the target reaction.
Thermochemistry
Thermochemistry is the branch of chemistry focused on the study of energy and heat associated with chemical reactions and changes of state. An essential part of thermochemistry is understanding how energy, particularly heat, is absorbed or released during chemical processes.

In our exercise, thermochemistry is applied to evaluate the energy dynamics of a chemical reaction involving aluminum and iron(III) oxide. One of the main tools in thermochemistry is the concept of enthalpy. Observing the change in enthalpy allows chemists to predict whether a reaction will release or absorb heat. Knowing whether a reaction is exothermic (releases heat) or endothermic (absorbs heat) helps us understand both practical and theoretical aspects of chemical reactions.

A firm grasp of thermochemistry principles assists in various applications, such as designing combustion engines or optimizing industrial chemical processes, where energy efficiency is crucial.

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Most popular questions from this chapter

A sheet of gold weighing \(10.0 \mathrm{~g}\) and at a temperature of \(18.0^{\circ} \mathrm{C}\) is placed flat on a sheet of iron weighing \(20.0 \mathrm{~g}\) and at a temperature of \(55.6^{\circ} \mathrm{C}\). What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

The standard enthalpy change for the following reaction is \(436.4 \mathrm{~kJ} / \mathrm{mol}\) : $$\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)$$ Calculate the standard enthalpy of formation of atomic hydrogen (H).

Ice at \(0^{\circ} \mathrm{C}\) is placed in a Styrofoam cup containing \(361 \mathrm{~g}\) of a soft drink at \(23^{\circ} \mathrm{C}\). The specific heat of the drink is about the same as that of water. Some ice remains after the ice and soft drink reach an equilibrium temperature of \(0^{\circ} \mathrm{C}\). Determine the mass of ice that has melted. Ignore the heat capacity of the cup.

What is meant by the standard enthalpy of a reaction?

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution? Assume the specific heat of the solution is the same as that for pure water.

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