\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}

Short Answer

Expert verified
The bond energy per mole for the C-H bond in CH4 is 1660 kJ/mol.

Step by step solution

01

Identify Known Energies

Start by identifying the known energies: The heat of dissociation of \( \mathrm{H}_{2} \) is 435 kJ/mol, the energy required for the sublimation of carbon (from \( \mathrm{C}_{(\mathrm{s})} \) to \( \mathrm{C}_{(\mathrm{g})} \)) is 720 kJ/mol, and the heat of formation (\( \Delta H_{\mathrm{f}} \)) for \( \mathrm{CH}_{4} \) is -75 kJ/mol.
02

Calculate Heat of Formation of 2 Mols of Hydrogen Atoms

Since one mole of \( \mathrm{H}_{2} \) gas dissociates into two moles of hydrogen atoms, we need to double the heat of dissociation to account for both atoms. Thus, the energy for two moles of \( \mathrm{H} \) atoms is \( 2 \times 435 \) kJ/mol.
03

Sum the Energies Required to Break Bonds

To determine the total energy required to break the bonds in one mole of methane to give one mole of carbon atoms and four moles of hydrogen atoms, sum the energy for the sublimation of carbon and the dissociation energy for hydrogen atoms. The total bond energy per mol of \( \mathrm{CH}_{4} \) is the sum of 720 kJ/mol for carbon and \( 2 \times 435 \) kJ/mol for hydrogen.
04

Subtract Heat of Formation to Find C-H Bond Energy

The bond energy per mole of the \( \mathrm{C}-\mathrm{H} \) bond is therefore obtained by subtracting the heat of formation of \( \mathrm{CH}_{4} \) from the total bond energy. Calculate this value using the equation: bond energy = total bond energy per mol of \( \mathrm{CH}_{4} \) - heat of formation.

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.

Heat of Dissociation
When we talk about the heat of dissociation, we are referring to the amount of energy required to break a particular chemical bond in a molecule to form separate atoms. This is a fundamental concept in thermochemistry and crucial for understanding how molecules interact. In the context of the given exercise, the heat of dissociation pertains to the energy needed to split molecular hydrogen (H_{2}) into individual hydrogen atoms. Calculating this heat for diatomic molecules like H_{2}, which have only one type of bond, is relatively straightforward. This information is pivotal when we examine more complex molecules and need to understand the individual bond energies to predict reactions' behavior.

For instance, if the heat of dissociation for H_{2} is given as 435 kJ/mol, as in our example, this value indicates the energy required to break the bonds in one mole of H_{2} molecules. In thermochemical calculations such as those for IIT-JEE exams, students need to pay attention to the stoichiometry of the reaction, where calculating the energy for forming or breaking multiple bonds becomes necessary.
Enthalpy of Formation
Enthalpy of formation, denoted as Delta H_{f}, is another key concept in thermochemistry that represents the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It's a measure of the energy change during the formation of a substance from its constituent elements. A negative Delta H_{f} is indicative of an exothermic reaction, where energy is released when the compound forms. Conversely, a positive Delta H_{f} would signify an endothermic process, requiring energy input to occur.

In our exercise, the enthalpy of formation for methane (CH_{4}) is provided as -75 kJ/mol. This implies that forming methane from its elements - carbon and hydrogen - releases energy, making the reaction exothermic. This energy must be taken into consideration when calculating bond energies within a molecule because it affects the total energy needed to break the molecule down into its atomic constituents.
Sublimation Energy
Sublimation energy is the amount of energy required to convert a substance from the solid phase directly into the gas phase without passing through the liquid phase. This is an important concept in thermochemistry, especially when studying elements like carbon, which sublimate at standard pressure conditions. The sublimation energy plays a pivotal role when calculating the overall energy changes in reactions involving elements that sublimate.

In the given problem, the exercise states that the sublimation energy of carbon, converting from solid (C_{(s)}) to gas (C_{(g)}), is 720 kJ/mol. This amount of energy is significant when calculating the bond energy for the C-H bond in CH_{4} because it represents the energy investment required to free the carbon atoms into a gaseous state where they can then bond with hydrogen atoms.
Thermochemistry for IIT-JEE
Thermochemistry is a branch of physical chemistry that involves the study of heat evolved or absorbed in chemical reactions. It's a key topic for competitive exams like IIT-JEE, where a deep understanding of energetic aspects of chemical reactions is essential. Mastering concepts such as the ones discussed here - heat of dissociation, enthalpy of formation, and sublimation energy - is crucial for success in IIT-JEE and other similar exams.

As it pertains to thermochemistry questions in IIT-JEE, students are often required to perform calculations involving these concepts to predict the possible outcomes of reactions or to determine the feasibility of a reaction under given conditions. It involves a careful balance and understanding of various types of energies, including bond energies and the energy changes associated with phase changes. To excel in thermochemistry, one must practice manipulating these energy values and applying the laws of thermodynamics in a wide range of chemical scenarios.

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

\(1.00\) 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 \(\mathrm{mol}^{-1}\).

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.

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.

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 \(\mathrm{H}_{2} \mathrm{O}\).

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