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Use the values of \(\Delta H_{\text {rxn }}^{\circ}\) given for the equations $$ \begin{aligned} 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-824.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ \(3 \mathrm{Fe}(s)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)\) $$ \Delta H_{\mathrm{rxn}}^{\circ}=-1118.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} $$ to calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the equation $$ 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\frac{1}{2} \mathrm{O}_{2}(g) $$

Short Answer

Expert verified
The enthalpy change for the reaction is \(235.8 \mathrm{~kJ/mol}\).

Step by step solution

01

Write Known Reaction Equations

The given reactions with their enthalpy changes are: \(2 \mathrm{Fe}(s) + \frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) with \(\Delta H = -824.2 \mathrm{~kJ/mol}\) and \(3 \mathrm{Fe}(s) + 2 \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)\) with \(\Delta H = -1118.4 \mathrm{~kJ/mol}\).
02

Reverse First Reaction

Reverse the first equation so that \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) is a reactant:\[\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 2 \mathrm{Fe}(s) + \frac{3}{2} \mathrm{O}_{2}(g)\] This changes the sign of \(\Delta H\) to \(+824.2 \mathrm{~kJ/mol}\).
03

Scale First Reaction

Multiply the reversed equation by 3 so that it matches the coefficient of \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) in the target equation:\[3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 6 \mathrm{Fe}(s) + \frac{9}{2} \mathrm{O}_{2}(g)\] The \(\Delta H\) for this scaled reaction is \(3 \times 824.2 \mathrm{~kJ/mol} = 2472.6 \mathrm{~kJ/mol}\).
04

Scale Second Reaction

Multiply the second equation by 2:\[6 \mathrm{Fe}(s) + 4 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)\] The \(\Delta H\) for this reaction becomes \(2 \times -1118.4 \mathrm{~kJ/mol} = -2236.8 \mathrm{~kJ/mol}\).
05

Combine Scaled Reactions

Add the scaled equations from Step 3 and Step 4 together:\[3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s) + \frac{1}{2} \mathrm{O}_{2}(g)\] This is the target reaction. The total enthalpy change is:\[\Delta H = 2472.6 \mathrm{~kJ/mol} + (-2236.8 \mathrm{~kJ/mol}) = 235.8 \mathrm{~kJ/mol}\]
06

Conclusion

The enthalpy change for the desired reaction \(3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s) + \frac{1}{2} \mathrm{O}_{2}(g)\) is \(\Delta H = 235.8 \mathrm{~kJ/mol}\).

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted as \(\Delta H\), is a key concept in thermochemistry that helps us understand the heat involved in chemical reactions. When a reaction occurs, energy can either be absorbed or released. This energy change is the enthalpy change of the reaction.

For exothermic reactions, energy is released, and \(\Delta H\) is negative. On the other hand, endothermic reactions absorb energy, resulting in a positive \(\Delta H\). This is important for predicting how a reaction will behave.

Let's take an example from the exercise: \( 2 \text{Fe}(s) + \frac{3}{2} \text{O}_{2}(g) \rightarrow \text{Fe}_{2} \text{O}_{3}(s)\) with \(\Delta H = -824.2 \text{kJ/mol}\). This negative sign indicates exothermic behavior. Conversely, reversing this reaction changes the sign, highlighting the principle of enthalpy's dependence on reaction direction.
Chemical Equations
Chemical equations are a foundation of understanding reactions. They describe what happens in a reaction, showing reactants and products with their respective phases.

Let's remember that chemical equations must be balanced. That means the number of atoms for each element should be the same on both sides of the equation. This reflects the conservation of mass principle. For example, the exercise presents the equation: \(3 \text{Fe}_{2} \text{O}_{3}(s) \rightarrow 2 \text{Fe}_{3} \text{O}_{4}(s) + \frac{1}{2} \text{O}_{2}(g)\). Balancing equations is crucial as it directly influences stoichiometry and reaction calculations.

In performing calculations with Hess's Law, like scaling and reversing equations, it's essential to adhere to these balanced relationships, as each coefficient plays a role in not only balancing but also calculating enthalpy changes.
Thermochemistry
Thermochemistry is the study of the heat involved in chemical reactions. It merges chemistry with thermodynamics to understand energy changes that dictate reaction spontaneity.

Hess's Law, a pivotal part of thermochemistry, states that the total enthalpy change for a reaction is the same, regardless of the number of steps or its pathway. This allows for the calculation of impossible-to-measure reactions through known equations and their enthalpy values.

For instance, to solve the problem in this exercise, we broke down the reaction into two known reactions, adjusted their coefficients, and combined them to find the enthalpy change for the unknown reaction. Thermochemistry equips us with the tools to not only calculate such values but also to understand the flow of energy, helping us predict and control chemical changes.

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Most popular questions from this chapter

A researcher studying the nutritional value of a new candy places a \(5.00\) -gram sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is \(2.89^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter is \(38.70 \mathrm{~kJ} \cdot \mathrm{K}^{-1}\), how many nutritional Calories are there per gram of the candy? Assume \(\Delta H_{\mathrm{rxn}}^{\circ} \approx \Delta U_{\mathrm{rxn}}^{\circ}\)

According to a U.S. Nutrition Facts label on a package of potato chips, a \(10.5\) -ounce package provides 130 nutritional Calories perserving. One serving is listed by the manufacturer as being \(1.125\) ounces. What is the total nutritional Calorie content of the package? In most other countries food labels list the total energy content of the package in kilojoules. How would the energy content of this package be labeled in other countries?

How much work (in joules) does a system do if its volume increases from 10 liters to 25 liters against a constant pressure of \(3.5 \mathrm{~atm} ?\)

Ammonium nitrate is an explosive at high temperatures and can decompose according to the equation $$ 2 \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow 2 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) $$ A \(1.00\) -gram sample of \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)\) is detonated in a bomb calorimeter with a total heat capacity of \(4.92 \mathrm{~kJ} \cdot \mathrm{K}^{-1} .\) The temperature increase is \(0.300 \mathrm{~K} .\) Calculate the energy evolved as heat (in kilojoules) for the decomposition of \(1.00\) kilogram of ammonium nitrate under these conditions.

Given that $$ \begin{array}{ll} \mathrm{Xe}(g)+\mathrm{F}_{2}(g) \rightarrow \mathrm{XeF}_{2}(s) & \Delta H_{\mathrm{rxn}}^{\circ}=-123 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\ \mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightarrow \mathrm{XeF}_{4}(s) & \Delta H_{\mathrm{rxn}}^{\circ}=-262 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{array} $$ calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the equation $$ \mathrm{XeF}_{2}(s)+\mathrm{F}_{2}(g) \rightarrow \mathrm{XeF}_{4}(s) $$

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