The standard enthalpy of formation of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is \(-65 \mathrm{kcal}\) \(\mathrm{mol}^{-1}\) and \(-197\) kcal mol \(^{-1}\) respectively. A mixture of two oxides contains \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the mole ratio \(2: 1 .\) If by oxidation, it is changed into a \(1: 2\) mole ratio mixture, how much of thermal energy will be released per mol of initial mixture.

Short Answer

Expert verified
-51 kcal/mol thermal energy will be released per mol of initial mixture.

Step by step solution

01

Identify Given Information

The standard enthalpy of formation (ΔH_formation) of FeO is -65 kcal/mol and for Fe_2O_3 is -197 kcal/mol. The initial mixture has a mole ratio of FeO to Fe_2O_3 of 2:1 and the final mixture after oxidation has a mole ratio of 1:2.
02

Determine Initial and Final Enthalpies

Calculate the initial and final enthalpies of the mixtures. For the initial mixture (2 moles of FeO and 1 mole of Fe_2O_3), the enthalpy is: 2(-65 kcal/mol) + 1(-197 kcal/mol). For the final mixture (1 mole of FeO and 2 moles of Fe_2O_3), the enthalpy is: 1(-65 kcal/mol) + 2(-197 kcal/mol).
03

Calculate the Change in Enthalpy

Subtract the final mixture's enthalpy from the initial mixture's enthalpy to find the thermal energy released during the oxidation process per mole of initial mixture.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Enthalpy Change Calculation
Understanding enthalpy change calculations is critical when studying reactions in chemistry. Enthalpy, denoted by the symbol H, is the total heat content of a system. It's a thermodynamic quantity that's difficult to measure directly, but changes in enthalpy can be calculated based on the heat absorbed or released during a chemical reaction at constant pressure.

For the given exercise, we need to calculate the enthalpy change that occurs when a mixture of iron(II) oxide (FeO) and iron(III) oxide (Fe2O3) undergoes oxidation, changing its mole ratio. We're provided with the standard enthalpies of formation for both oxides, which represent the enthalpy change when one mole of a compound is formed from its elements in their standard states.

To calculate the enthalpy change for the reaction, we need the enthalpies of the initial and final mixtures. By applying Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how many steps it involves, we can use the standard enthalpies of formation to find the total enthalpy of the mixture before and after the reaction:
  • Initial mixture enthalpy: 2(-65 kcal/mol) + 1(-197 kcal/mol)
  • Final mixture enthalpy: 1(-65 kcal/mol) + 2(-197 kcal/mol)
The change in enthalpy (ablaH) is then found by subtracting the final enthalpy from the initial enthalpy.
Chemical Thermodynamics
Chemical thermodynamics deals with the energy changes associated with chemical reactions and the direction in which these reactions occur. The enthalpy change calculation discussed in the previous section is just one aspect of this vast field. Essentially, thermodynamics tells us whether a process is likely to happen and provides insights into how much energy might be involved.

From the thermodynamic perspective, our mixture's oxidation process illustrates an exothermic reaction, where thermal energy is released into the surroundings—indicated by the negative sign in standard enthalpies of formation. This release of energy tends to make the reaction spontaneous, a principle dictated by Gibbs free energy which combines enthalpy, temperature, and entropy to predict the favorability of reactions.

The concept of standard enthalpy of formation, used in the given exercise, is a key concept in thermodynamics that helps us compare the energy required to break bonds in reactants with the energy released upon bond formation in products. These values are critical for scientists and engineers designing processes that take advantage of exothermic reactions, like combustion engines or chemical heaters.
Mole Ratio in Reactions
The concept of mole ratio in reactions is foundational in stoichiometry, which is the study of quantitative relationships in chemical reactions. Mole ratio defines the proportional relationship between reactants and products involved in a chemical reaction and is derived from the balanced chemical equation.

In our textbook exercise, the mole ratio changes due to the oxidation process. Initially, the ratio is 2:1 for FeO to Fe2O3. After oxidation, the ratio becomes 1:2. This change directly affects the enthalpy calculation, as the number of moles dictates the total enthalpy of each substance, which we use to find the thermal energy released. It's crucial to understand that even without knowing the full balanced chemical equations, the mole ratio can provide us with enough information to calculate the enthalpy changes for the oxidation of the mixture.

With this in mind, students can improve their understanding of this exercise by ensuring they are clear on the concept of mole ratios and how they apply to the conservation of mass and energy within a reaction. It's the link between the microscale (atoms and molecules) and the macroscale (grams and liters) and is central to correctly predicting the outcomes of chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The heat of solution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in water was determined by measuring the amount of electrical work needed to compensate for the cooling which would otherwise óccur when the salt dissolves. After the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) was added to the water, electrical energy was provided by passage of current through a resistance coil until the temperature of the solution reached the value it had prior to the addition of salt. In a typical experiment, \(4.4 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) was added to \(200 \mathrm{~g}\) water. A current of \(0.75\) ampere was provided through the heater coil, and the voltage across the terminals was \(6.0 \mathrm{~V}\). The current was applied for \(5.2\) minute. Calculate \(\Delta H\) for the solution of \(1.0\) mole \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in enough water to give same concentration as was attained in the above experiment.

The integral enthalpy of solution in \(\mathrm{kJ}\) of one mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(n\) mole of water is given by: $$ \Delta H_{s}=\frac{75.6 \times n}{n+1.8} $$ Calculate \(\Delta H\) for the following process: (a) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(2 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\). (b) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in 7 mole of \(\mathrm{H}_{2} \mathrm{O}\). (c) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in 5 mole of \(\mathrm{H}_{2} \mathrm{O}\). (d) solution (a) dissolved in 5 mole of \(\mathrm{H}_{2} \mathrm{O}\). (e) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in excess of \(\mathrm{H}_{2} \mathrm{O}\).

\(1.00\) litre sample of a mixture of \(\mathrm{CH}_{4(\mathrm{~g})}\) and \(\mathrm{O}_{2(\mathrm{~g})}\) measured at \(25^{\circ} \mathrm{C}\) and 740 torr was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of \(1260 \mathrm{cal} / \mathrm{K}\). The complete combustion of the methane to \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) caused a temperature rise in the calorimeter of \(0.667 \mathrm{~K}\). What was the mole per cent of \(\mathrm{CH}_{4}\) in the original mixture? \(\Delta H_{\text {comb }}^{\circ}\left(\mathrm{CH}_{4}\right)=-215 \mathrm{k}\) cal \(\mathrm{mol}^{-1}\).

\begin{aligned} &\text { The heat of dissociation of } \mathrm{H}_{2} \text { is } 435 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { . If } \mathrm{C}_{(\mathrm{s})} \longrightarrow \mathrm{C}_{(\mathrm{g})} \text { ; }\\\ &\Delta H=720 \mathrm{~kJ} \mathrm{~mol}^{-1}, \text { calculate the bond energy per mol of the } \mathrm{C}-\mathrm{H} \text { bond }\\\ &\text { in } \mathrm{CH}_{4} \text { molecule. } \Delta H_{\mathrm{f}} \text { for } \mathrm{CH}_{4}=-75 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { . } \end{aligned}

The commercial production of water gas utilizes the reaction under standard conditions: \(\mathrm{C}+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}+\mathrm{CO}\). The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to \(\mathrm{CO}_{2}\). How many \(\mathrm{g}\) of carbon must be burnt to \(\mathrm{CO}_{2}\) to provide enough heat for the water gas conversion of \(100 \mathrm{~g}\) carbon? Neglect all heat losses to the environment. Also \(\Delta H_{\mathrm{f}}^{2}\) of \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) and \(\mathrm{CO}_{2}\) are \(-110.53\), \(-241.81\) and \(-393.51 \mathrm{~kJ} / \mathrm{mol}\) respectively.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free