Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 11

Which of the following substances would obey Trouton's rule most closely: HF, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\) (toluene), or \(\mathrm{CH}_{3} \mathrm{OH}\) (methanol)? Explain your reasoning.

Short Answer

Expert verified
Toluene (\(C_6H_5CH_3\)) would obey Trouton's rule most closely, as it is a non-polar molecule, while HF and methanol (\(CH_3OH\)) are polar.

Step by step solution

01

Identify the Polarity of HF

HF is a polar molecule. The difference in electronegativity between hydrogen and fluorine result in a polar bond, leading to a dipole in the molecule.
02

Identify the Polarity of C6H5CH3 (toluene)

Toluene (C6H5CH3) is a non-polar molecule. While the C-H bonds can be considered slightly polar, the symmetrical cyclic structure of the molecule results in the cancellation of dipole moments, leading to a non-polar molecule.
03

Identify the Polarity of CH3OH (methanol)

Methanol (CH3OH) is a polar molecule. It contains a -OH group, which is a polar functional group, resulting in a polar molecule.
04

Apply Trouton's Rule

Given that Trouton's Rule applies best to non-polar molecules, toluene (C6H5CH3), being a non-polar molecule, would obey Trouton's Rule most closely compared to HF and methanol (CH3OH), which are polar.

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.

Start your free trial

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 following standard Gibbs energy changes are given for \(25^{\circ} \mathrm{C}\) (1) \(\mathrm{SO}_{2}(\mathrm{g})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-246.4 \mathrm{kJ}\) (2) \(\mathrm{CS}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=-41.5 \mathrm{kJ}\) (3) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=+1.4 \mathrm{kJ}\) (4) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-28.6 \mathrm{kJ}\) Combine the preceding equations, as necessary, to obtain \(\Delta G^{\circ}\) values for the following reactions. (a) \(\operatorname{COS}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\begin{aligned} \mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \Delta G^{\circ}=? \end{aligned}\) (b) \(\cos (g)+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \quad \Delta G^{\circ}=?\) \(\left.+\quad \mathrm{H}_{\mathrm{O}} \mathrm{C}(\mathrm{d})=\mathrm{CO}_{-}^{\circ} \mathrm{G}\right)+\mathrm{H}_{-}^{-} \mathrm{S}(\mathrm{q})\) (c) \(\cos (\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=?\) Of reactions (a), (b), and (c), which is spontaneous in the forward direction when reactants and products are present in their standard states?

Use the following data to estimate the standard molar entropy of gaseous benzene at \(298.15 \mathrm{K} ;\) that is, \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \mathrm{atm})\right] .\) For \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{s}, 1 \mathrm{atm})\) at its melting point of \(5.53^{\circ} \mathrm{C}, S^{\circ}\) is \(128.82 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(9.866 \mathrm{kJ} \mathrm{mol}^{-1} .\) From the melting point to 298.15 K, the average heat capacity of liquid benzene is \(134.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The enthalpy of vaporization of \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at \(298.15 \mathrm{K}\) is \(33.85 \mathrm{kJ} \mathrm{mol}^{-1},\) and in the vapor- ization, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) is produced at a pressure of 95.13 Torr. Imagine that this vapor could be compressed to 1 atm pressure without condensing and while behaving as an ideal gas. Calculate \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \text { atm) }] .[ \text { Hint: Refer to }\right.\) Exercise \(88,\) and note the following: For infinitesimal quantities, \(d S=d q / d T ;\) for the compression of an ideal gas, \(d q=-d w ;\) and for pressure-volume work, \(d w=-P d V\).

What must be the temperature if the following reaction has \(\Delta G^{\circ}=-45.5 \mathrm{kJ}, \Delta H^{\circ}=-24.8 \mathrm{kJ},\) and \(\Delta S^{\circ}=15.2 \mathrm{JK}^{-1} ?\) $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})$$

Which of the following changes in a thermodynamic property would you expect to find for the reaction \(\mathrm{Br}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Br}(\mathrm{g})\) at all temperatures: \((\mathrm{a}) \Delta H<0\) (b) \(\Delta S>0 ;\) (c) \(\Delta G<0 ;\) (d) \(\Delta S<0 ?\) Explain.

Explain why (a) some exothermic reactions do not occur spontaneously, and (b) some reactions in which the entropy of the system increases do not occur spontaneously.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation
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