Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The heat of combustion of 1 mole of liquid decane to give carbon dioxide and liquid water is \(1620.1\) kcal The heat of vaporization of decane at \(25^{\circ}\) is \(11.7\) kcal \(\mathrm{mol}^{-1}\). Calculate the heat of combustion that would be observed for all the participants in the vapor phase.

Short Answer

Expert verified
The heat of combustion for gaseous decane is 1631.8 kcal.

Step by step solution

01

Understand the Problem

We need to find the heat of combustion for decane when all participants are in the vapor phase. We know the combustion reaction for liquid decane and the heat of vaporization. Our job is to adjust the heat of combustion to account for the phase change from liquid to vapor.
02

Define Given Data

The heat of combustion for 1 mole of liquid decane is given as \(1620.1\) kcal. The heat of vaporization for decane at \(25^{\circ}\) C is \(11.7\) kcal/mol. These values will be used to calculate the combustion heat for gaseous decane.
03

Calculate Heat for Phase Change

Since we need to convert liquid decane to gaseous decane, add the heat of vaporization to the heat of combustion. This is because extra energy is required to convert the liquid decane to vapor before combustion takes place.
04

Adjust Heat of Combustion

Add the heat of vaporization to the given heat of combustion: \(1620.1\) kcal + \(11.7\) kcal = \(1631.8\) kcal. This is the heat of combustion for gaseous decane where all participants are vapor.
05

Conclusion

The calculated heat of combustion considers the phase transition from liquid to gas for decane before undergoing combustion. This adjusted value represents the energy released when decane is in vapor form during combustion.

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 combustion
In thermochemistry, the heat of combustion refers to the energy released as heat when a substance undergoes complete combustion with oxygen. It is usually measured in units of kilocalories (kcal) or kilojoules (kJ) per mole of substance. When combusting decane, a hydrocarbon, the process involves breaking multiple carbon-hydrogen bonds to form carbon dioxide and water. This process releases a significant amount of energy.
One key point here is that the heat of combustion is typically calculated under standard conditions, which assume the products like carbon dioxide and water are in specified states – usually gaseous at certain temperatures. The given example talks about the combustion of liquid decane resulting in liquid water and gaseous carbon dioxide.
This process releases a lot of energy, and for liquid decane, it was specifically measured as 1620.1 kcal per mole. This indicates the energy obtained when decane burns fully, but it doesn't consider if decane were initially in the vapor phase, which requires additional energy adjustments.
Heat of vaporization
The heat of vaporization is the amount of energy required to transform a substance from a liquid into a gas or vapor. This transformation is a type of phase change, which is crucial in thermochemistry when considering energy calculations for reactions involving phase changes.
  • For decane, the provided heat of vaporization is 11.7 kcal per mole.
  • This means that it takes an additional 11.7 kcal for each mole of decane to transition from liquid to vapor before participating in a chemical reaction such as combustion.
Understanding this concept is essential because when substances like decane need to undergo a phase change before engaging in a reaction, this extra energy requirement must be added to the overall energy calculations. Hence, in exercises like our current example, heat of vaporization becomes a critical component in determining the complete energetic profile of the combustion process.
Phase change
A phase change refers to the transition of a substance from one state of matter to another, such as solid to liquid, liquid to gas, or vice versa. In thermochemistry, phase changes are significant because they involve energy absorption or release without changing the chemical identity of a substance.
In the example with decane, the phase change involves its transition from liquid to vapor prior to combustion. This transition requires energy, termed the heat of vaporization, as discussed earlier. When calculating energies in thermochemical reactions, accounting for phase changes is necessary to accurately reflect the conditions under which a reaction takes place.
  • Phase changes can either absorb energy (endothermic) or release energy (exothermic).
  • In the case of decane, moving from liquid to vapor requires an endothermic input of energy.
Thus, when considering the energy changes in a complete thermochemical process, including phase changes, ensures more accurate and realistic results, providing a better understanding of the nature and extent of the reactions involved.

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

A \(C-F\) bond energy can be computed from thermochemical studies of the vapor- phase reaction $$ \mathrm{CH}_{4}+4 \mathrm{~F}_{2} \rightarrow \mathrm{CF}_{4}+4 \mathrm{HF} \quad \Delta H^{0}=-460 \mathrm{kcal} $$ Show how the \(\Delta H^{0}\) value for this reaction may be used to calculate the energy of the \(\mathrm{C}-\mathrm{F}\) bond if all the other bond energies are known.

Knowing that the equilibrium constant \(K_{\text {eq for formation of nonane from solid carbon and hydrogen gas is }}\) \(4.7 \times 10^{-5}\), explain why liquid nonane does not forthwith decompose into its elements.

The entropy change \(\Delta S^{0}\) for the formation of chloroethane by chlorination of ethane is \(+0.5 \mathrm{e} . \mathrm{u}\), and for the formation of chloroethane by combination of hydrogen chloride with ethene \(\Delta S^{0}\) is \(-31\) e.u. Explain. $$ \begin{array}{ll} \mathrm{CH}_{3}-\mathrm{CH}_{3}+\mathrm{Cl}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{HCl} & \Delta S^{0}=+0.5 \mathrm{e.u} \\\ \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{HCl} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl} & \Delta S^{0}=-31 \mathrm{e.u} \end{array} $$

tert-Butyl hypobromite is a radical brominating agent that is similar to tert- butyl hypochlorite (Exercise 4 \(18^{*}\) ), but is less easily prepared than the hypochlorite. A good substitute, provided radical bromination is possible, is a mixture of \(\mathrm{BrCCl}_{3}\) and \(\mathrm{ce}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COCl}\) Thus, bromination of cyclohexene results if a high ratio of bromotrichloromethane to hypochlorite is used. O=C(O)CCC1CCCCC1Br Suggest how this reaction is initiated and propagated, and explain why it is necessary to have an excess of bromotrichloromethane.

The heat of combustion of cyclopropane, \(\left(\mathrm{CH}_{2}\right)_{3}\), to give carbon dioxide and liquid water is \(499.8 \mathrm{kcal} \mathrm{mol}^{-1}\). Show how this value, assuming normal \(\mathrm{C}-\mathrm{H}\) bond strengths, can be used to calculate the average \(\mathrm{C}-\mathrm{C}\) bond energy of cyclopropane.

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