/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Normal butane and isobutane have... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Normal butane and isobutane have boiling temperatures of \(-0.5\) and \(-12.3^{\circ} \mathrm{C}\) (31.1 and \(\left.9.9^{\circ} \mathrm{F}\right)\), respectively. Briefly explain this behavior on the basis of their molecular structures, as presented in Section 14.2.

Short Answer

Expert verified
Question: Explain the difference in boiling temperatures of normal butane and isobutane based on their molecular structures and intermolecular forces. Answer: Normal butane has a linear structure, providing a larger surface area for intermolecular interactions, while isobutane has a branched structure with less surface area. This results in stronger London dispersion forces in normal butane compared to isobutane, leading to a higher boiling temperature in normal butane (-0.5°C) than in isobutane (-12.3°C).

Step by step solution

01

Identify Molecular Structures

First, let's identify the molecular structures of normal butane and isobutane. These structures can be represented as follows: Normal butane: \(\mathrm{CH_3CH_2CH_2CH_3}\) Isobutane: \(\mathrm{(CH_3)_3CH}\) Both compounds are isomers of each other, meaning that they have the same molecular formula but different molecular structures.
02

Identify Intermolecular Forces

Now, let's identify the types of intermolecular forces present in both normal butane and isobutane. Since both compounds are nonpolar, their main intermolecular force is the London dispersion forces (also known as Van der Waals forces).
03

Compare Molecular Structures and Surface Area

It is important to notice that molecular structures play a crucial role in determining the strength of the London dispersion forces and consequently, the boiling temperature of the compounds. The London dispersion forces are stronger when molecules have larger surface areas, as it allows for greater contact and interaction between neighboring molecules. Normal butane has a linear structure that provides a larger surface area for interactions between neighboring molecules. On the other hand, isobutane has a branched structure, which gives less surface area for interactions with neighboring molecules.
04

Relate Boiling Temperatures to Intermolecular Forces

As mentioned before, stronger intermolecular forces lead to a higher boiling temperature because it requires more energy to break these forces. In this case, the London dispersion forces are comparatively stronger in normal butane due to larger surface area than in isobutane. Therefore, normal butane has a higher boiling temperature than isobutane. Normal butane boiling temperature: \(-0.5^{\circ} \mathrm{C}\) Isobutane boiling temperature: \(-12.3^{\circ} \mathrm{C}\) In conclusion, the difference in boiling temperatures of normal butane and isobutane can be explained by their molecular structures, which affect the surface area and consequently the strength of the London dispersion forces in these compounds.

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 91Ó°ÊÓ!

Key Concepts

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

Molecular Structures
Molecular structures are essential in understanding the properties and behavior of molecules like normal butane and isobutane. These two molecules are structural isomers, which means they share the same molecular formula, \((C_4H_{10})\), but differ in the arrangement of their atoms.

Normal butane has a **linear structure** (\(CH_3CH_2CH_2CH_3\)), allowing the chain to stretch out in a straight line. This shape ensures a larger surface area compared to its branched counterpart.

Isobutane, represented as \((CH_3)_3CH\), exhibits a **branched structure**. This structure is more compact and spherical, reducing the surface area available for molecular interaction.
  • **Linear structures**: Provide more surface area, allowing more contact between molecules.
  • **Branched structures**: Have a compact form, limiting interactions between molecules.
The differences in these molecular structures are significant because they influence how the molecules interact through intermolecular forces, leading to variations in physical properties such as boiling temperatures.
London Dispersion Forces
London Dispersion Forces (LDFs), also known as Van der Waals forces, are the primary intermolecular forces in nonpolar molecules like normal butane and isobutane.

These forces are **temporary attractions** that arise when the motion of electrons creates an instantaneous dipole in a molecule, inducing a dipole in neighboring molecules.
This transient phenomenon is more effective in **molecules with larger surface areas**.
  • **Temporary attractions**: They happen due to fluctuations in the electron distribution in atoms and molecules.
  • **Influence of surface area**: A larger surface area allows greater interaction and stronger LDFs.
In the context of normal butane and isobutane, the strength of the London Dispersion Forces is determined by their surface area. The linear structure of normal butane allows for **greater surface interactions** compared to the compact branched structure of isobutane. Consequently, the differences in these forces contribute significantly to their boiling temperature differences.
Boiling Temperature
The boiling temperature of a substance is directly influenced by the strength of intermolecular forces present within it. For nonpolar molecules like normal butane and isobutane, **London Dispersion Forces** are the main contributors.

Normal butane, with its linear molecular structure, presents a larger surface area conducive to stronger London Dispersion Forces. This results in a relatively higher boiling temperature of \(-0.5^{\circ} \mathrm{C}\).

In contrast, **isobutane's branched structure** limits such interactions, resulting in weaker forces and a lower boiling temperature of \(-12.3^{\circ} \mathrm{C}\).
  • **Stronger forces** lead to higher boiling points because more energy is required to separate the molecules.
  • **Weaker forces** result in lower boiling points as less energy is needed to break the interactions.
Understanding the relationship between molecular structure and boiling temperature clarifies why despite sharing the same molecular formula, normal butane and isobutane have such differ in their boiling points.

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

Name the following polymer(s) that would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?

For each of the following pairs of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if his is possible, note which has the higher ensile modulus and then cite the reason(s) or your choice; and (3) if it is not possible o decide, then state why. a) Random acrylonitrile-butadiene copolyner with \(10 \%\) of possible sites crosslinked; ilternating acrylonitrile-butadiene copolymer with \(5 \%\) of possible sites crosslinked (b) Branched and syndiotactic polypropylene with a degree of polymerization of 5000 ; linear and isotactic polypropylene with a degree of polymerization of 3000 (c) Branched polyethylene with a numberaverage molecular weight of \(250,000 \mathrm{~g} / \mathrm{mol}\); linear and isotactic poly(vinyl chloride) with a number-average molecular weight of \(200,000 \mathrm{~g} / \mathrm{mol}\)

Briefly explain how each of the following influences the tensile or yield strength of a semicrystalline polymer and why: (a) Molecular weight (b) Degree of crystallinity (c) Deformation by drawing (d) Annealing of an undeformed material

List the two molecular characteristics that are essential for elastomers.

For the nylon polymer, whose stress-strain behavior may be observed in the "Tensile Tests" module of Virtual Materials Science and Engineering \((V M S E)\), determine the following: (a) the yield strength, and (b) the approximate ductility, in percent elongation. How do these values compare with those for the nylon material presented in Table 15.1?

See all solutions

Recommended explanations on Physics 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.