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The enthalpies of formation of \(\mathrm{FeO}(\mathrm{s})\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})\) are \(-65.0\) and \(-197.0 \mathrm{kcal} /\) mol, respectively. A mixture of the two oxides contains \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the mole ratio \(2: 1 .\) If by oxidation it is changed in to a \(1: 2\) mole ratio mixture, how much of thermal energy will be released per mole of the initial mixture? (a) \(13.4 \mathrm{kcal}\) (b) \(67 \mathrm{kcal}\) (c) \(47.2 \mathrm{kcal}\) (d) 81 kcal

Short Answer

Expert verified
The amount of thermal energy released per mole of the initial mixture is approximately 47.2 kcal, which corresponds to option (c).

Step by step solution

01

Establish the given enthalpies and mole ratios

The enthalpies of formation for FeO(s) and Fe_{2}O_{3}(s) are given as -65.0 kcal/mol and -197.0 kcal/mol, respectively. Initially, the mole ratio of FeO to Fe_{2}O_{3} is 2:1, and after oxidation, it becomes 1:2.
02

Determine the enthalpy change for the initial mixture

Calculate the total enthalpy of the initial mixture which contains 2 moles of FeO and 1 mole of Fe_{2}O_{3}. Total enthalpy = (2 moles * -65.0 kcal/mol) + (1 mole * -197.0 kcal/mol) = -2 * 65.0 kcal - 197.0 kcal = -327.0 kcal
03

Determine the enthalpy change for the final mixture

Calculate the total enthalpy of the final mixture, which contains 1 mole of FeO and 2 moles of Fe_{2}O_{3}.Total enthalpy = (1 mole * -65.0 kcal/mol) + (2 moles * -197.0 kcal/mol) = -65.0 kcal - 2 * 197.0 kcal = -459.0 kcal
04

Calculate the thermal energy released

Find the difference in enthalpy between the final and initial mixtures to determine the amount of thermal energy released.Energy released = Final enthalpy - Initial enthalpy = -459.0 kcal - (-327.0 kcal) = -459.0 kcal + 327.0 kcal = -132.0 kcal
05

Calculate the energy released per mole of initial mixture

Since the initial mixture is based on 3 moles total (2 moles of FeO and 1 mole of Fe_{2}O_{3}), we need to divide the total energy by 3 to find the energy per mole.Energy per mole = Total energy released / Total moles of initial mixture = -132.0 kcal / 3 moles = -44.0 kcal/mole
06

Analyze the options and give the final answer

Since -44.0 kcal/mole is not an option and considering energy release is an exothermic process, we should take the absolute value to represent the amount of energy released, which is 44.0 kcal/mole. The closest option to this value is (c) 47.2 kcal.

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Key Concepts

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

Enthalpies of Formation
Enthalpies of formation refer to the heat change that occurs when one mole of a compound is formed from its elements in their standard states. It's a fundamental concept in chemical thermodynamics, allowing chemists to predict the energy changes involved when substances react to form new compounds.

For instance, in our exercise, we have iron(II) oxide (FeO) and iron(III) oxide (Fe2O3) with their respective enthalpies of formation at -65.0 kcal/mol and -197.0 kcal/mol. A negative value indicates that the formation of the compound from its elements releases energy, thus being an exothermic process. By understanding these values, we can calculate the total enthalpy of a mixture of these compounds, which is essential for determining how much energy is involved in a reaction.
Chemical Thermodynamics
Chemical thermodynamics deals with the relationship between heat and other forms of energy in the context of chemical and physical processes. It's a cornerstone of understanding how energy is transferred and conserved within chemical systems.

In our example, we look at the transfer of energy when a mixture of iron oxides changes its composition due to oxidation. Thermodynamics allows us to interpret the enthalpies of formation and use them to determine the total enthalpy before and after the reaction. This enables us to determine the thermal energy released during the process, which is a concept closely tied to the first and second laws of thermodynamics—conservation of energy and the increase in entropy, respectively.
Exothermic Reactions
Exothermic reactions are reactions that release heat to the surroundings. They are characterized by a negative enthalpy change since the energy needed to break bonds in the reactants is less than the energy released when new bonds are formed in the products.

Considering the exercise, the oxidation that changes the mole ratio of FeO to Fe2O3 is exothermic, resulting in the release of thermal energy. In thermodynamics terms, the total enthalpy of the mixture decreases, illustrating that more energy is being released than absorbed. Our calculation approach must reflect this by showing a negative change in enthalpy, whose absolute value indicates the amount of energy given off to the environment.
Stoichiometry
Stoichiometry is the quantitative relationship between the amounts of reactants and products in a chemical reaction. It's like the 'recipe' for a reaction that tells us how much of each substance is needed and produced.

In the given exercise, stoichiometry is used to account for the changing amounts of FeO and Fe2O3. Initially, we have a 2:1 mole ratio of FeO to Fe2O3, which later changes to 1:2 following the reaction. To accurately calculate the energy change, we must consider these stoichiometric ratios to determine how much thermal energy is released per mole of the initial mixture. This involves dividing the total energy change by the total number of moles present at the beginning of the reaction.

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Most popular questions from this chapter

Study the following thermochemical data: \(\mathrm{S}+\mathrm{O}_{2} \rightarrow \mathrm{SO}_{2} ; \quad \Delta H=-298.2 \mathrm{~kJ}\) \(\mathrm{SO}_{2}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{SO}_{3} ; \quad \Delta H=-98.2 \mathrm{~kJ}\) \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4} ; \quad \Delta H=-130.2 \mathrm{~kJ}\) \(\mathrm{H}_{2}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} ; \quad \Delta H=-287.3 \mathrm{~kJ}\) The enthalpy of formation of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) at \(298 \mathrm{~K}\) will be (a) \(-433.7 \mathrm{k} \mathrm{J}\) (b) \(-650.3 \mathrm{~kJ}\) (c) \(+320.5 \mathrm{~kJ}\) (d) \(-813.9 \mathrm{~kJ}\)

The enthalpy of formation of ammonia gas is \(-46.0 \mathrm{~kJ} / \mathrm{mol}\). The enthalpy change for the reaction: \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) is (a) \(46.0 \mathrm{~kJ}\) (b) \(92.0 \mathrm{~kJ}\) (c) \(23.0 \mathrm{~kJ}\) (d) \(-92.0 \mathrm{~kJ}\)

Calculate the standard free energy of the reaction at \(27^{\circ} \mathrm{C}\) for the combustion of methane using the given data: \(\mathrm{CH}_{4}(\mathrm{~g})\) \(+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) $$\begin{array}{lcccc} \hline \text { Species } & \mathbf{C H}_{4}(\mathrm{~g}) & \mathrm{O}_{2}(\mathrm{~g}) & \mathrm{CO}_{2}(\mathrm{~g}) & \mathbf{H}_{2} \mathrm{O}(\mathrm{l}) \\ \hline \Delta_{\mathrm{f}} \boldsymbol{H}^{\circ} /(\mathrm{kJ} & -74.5 & 0 & -393.5 & -286.0 \\ \left.\mathrm{~mol}^{-1}\right) & & & & \\ \boldsymbol{S}^{\circ} /\left(\mathrm{JK}^{-1}\right. & 186 & 205 & 212 & 70 \\\ \left.\mathbf{m o l}^{-1}\right) & & & & \\ \hline \end{array}$$ (a) \(-891.0 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-240 \mathrm{~kJ} / \mathrm{mol}\) (c) \(-819 \mathrm{~kJ} / \mathrm{mol}\) (d) \(-963 \mathrm{~kJ} / \mathrm{mol}\)

The value of \(\Delta_{\mathrm{f}} H^{\circ}\) of \(\mathrm{U}_{3} \mathrm{O}_{8}(\mathrm{~s})\) is \(-853.5 \mathrm{~kJ}\) \(\mathrm{mol}^{-1} . \Delta H^{\circ}\) for the reaction: \(3 \mathrm{UO}_{2}(\mathrm{~s})\) \(+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{U}_{3} \mathrm{O}_{8}(\mathrm{~s})\), is \(-76.00 \mathrm{~kJ} .\) The value of \(\Delta_{\mathrm{f}} H^{\circ}\) of \(\mathrm{UO}_{2}(\mathrm{~s})\) is (a) \(-259.17 \mathrm{~kJ}\) (b) \(-310.17 \mathrm{~kJ}\) (c) \(+259.17 \mathrm{~kJ}\) (d) \(930.51 \mathrm{~kJ}\).

The intermediate \(\mathrm{SiH}_{2}\) is formed in the thermal decomposition of silicon hydrides. Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{SiH}_{2}\) from the following reactions: \(\mathrm{Si}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SiH}_{4}(\mathrm{~g})\) \(\Delta H^{\circ}=-11.7 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{SiH}_{4}(\mathrm{~g}) \rightarrow \mathrm{SiH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) \(\Delta H^{\circ}=+239.7 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta H_{\mathrm{f}}^{\circ}, \mathrm{Si}_{2} \mathrm{H}_{6}(\mathrm{~g})=+80.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(353 \mathrm{~kJ} / \mathrm{mol}\) (b) \(321 \mathrm{~kJ} / \mathrm{mol}\) (c) \(198 \mathrm{~kJ} / \mathrm{mol}\) (d) \(274 \mathrm{~kJ} / \mathrm{mol}\)

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