The normal boiling point of \(n\) -octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\( is \)125^{\circ} \mathrm{C}$. (a) Is the condensation of gaseous \(n\) -octane to liquid \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the boiling of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and gaseous \(n\) -octane are in equilibrium? Explain.

Short Answer

Expert verified
(a) The condensation of gaseous n-octane to liquid n-octane is an exothermic process. (b) The temperature range for spontaneous boiling of n-octane is above \(125 ^{\circ}{\rm C}\). (c) The temperature range for nonspontaneous boiling of n-octane is below \(125 ^{\circ}{\rm C}\). (d) Liquid n-octane and gaseous n-octane are in equilibrium at the normal boiling point of \(125 ^{\circ}{\rm C}\).

Step by step solution

01

(a) Identify the process type

To determine whether the condensation of gaseous n-octane to liquid n-octane is endothermic or exothermic, we consider energy changes. In an endothermic process, energy is absorbed from the surroundings, while in an exothermic process, energy is released to the surroundings. When a substance condenses, the molecules change from a less ordered, high-energy (gaseous) state to a more ordered, low-energy (liquid) state, implying a release of energy to the surroundings. Therefore, the condensation of gaseous n-octane to liquid n-octane is an exothermic process.
02

(b) Determine the temperature range for spontaneous boiling

Boiling occurs when the liquid phase and gaseous phase are in equilibrium at atmospheric pressure. In the case of n-octane, the normal boiling point is given as \(125 ^{\circ}{\rm C}\). At temperatures above this point, it is energetically favorable for the substance to be in its gaseous state, which means the boiling is spontaneous. Therefore, the temperature range for spontaneous boiling of n-octane is above \(125 ^{\circ}{\rm C}\).
03

(c) Determine the temperature range for nonspontaneous boiling

On the other hand, if the substance is below its normal boiling point, it will tend to remain in its condensed (liquid) state, as it would not have enough energy to transition to its gaseous state. In this condition, the boiling process is nonspontaneous. Therefore, the temperature range for nonspontaneous boiling of n-octane is below \(125 ^{\circ}{\rm C}\).
04

(d) Identify the equilibrium temperature

The equilibrium temperature between the liquid and gaseous phases is the normal boiling point, at which the vapor pressure of the liquid is equal to the atmospheric pressure. For n-octane, this occurs at \(125 ^{\circ}{\rm C}\). So, yes, there is a temperature at which liquid n-octane and gaseous n-octane are in equilibrium, and that temperature is the normal boiling point of \(125 ^{\circ}{\rm C}\).

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!

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 conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$, is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C\), calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S)\) ? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, \((\mathbf{c})\) a solid sublimes?

Use data from Appendix \(C\) to calculate the equilibrium constant, \(K,\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) (b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (c) $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta H\) zero for the process?

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(I_{2}\). (c) Which of the values in part (b) is closer to the value you obtained in part (a)?

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