Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 29

Look up and compare the normal boiling points and normal melting points of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{S}\) . Based on these physical properties, which substance has stronger intermolecular forces? What kinds of intermolecular forces exist for each molecule?

Short Answer

Expert verified
The normal boiling point of \(\mathrm{H}_2\mathrm{O}\) is 100 °C (373 K) while \(\mathrm{H}_2\mathrm{S}\) is -60.4 °C (212.8 K). The normal melting point of \(\mathrm{H}_2\mathrm{O}\) is 0 °C (273 K) while \(\mathrm{H}_2\mathrm{S}\) is -85.5 °C (187.7 K). As \(\mathrm{H}_2\mathrm{O}\) has higher boiling and melting points, it has stronger intermolecular forces than \(\mathrm{H}_2\mathrm{S}\). \(\mathrm{H}_2\mathrm{O}\) has dipole-dipole interactions and hydrogen bonding, while \(\mathrm{H}_2\mathrm{S}\) has weaker dipole-dipole interactions and no hydrogen bonding, resulting in the difference in intermolecular force strengths.

Step by step solution

01

Look up the normal boiling and melting points

For this step, we need to search for the normal boiling and melting points of \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{H}_2\mathrm{S}\). - The normal boiling point of \(\mathrm{H}_2\mathrm{O}\) is 100 °C (373 K) and its normal melting point is 0 °C (273 K). - The normal boiling point of \(\mathrm{H}_2\mathrm{S}\) is -60.4 °C (212.8 K) and its normal melting point is -85.5 °C (187.7 K).
02

Compare boiling and melting points

Based on the values from step 1, we can compare the boiling and melting points between the two substances. \(\mathrm{H}_2\mathrm{O}\) has a higher boiling and melting point compared to \(\mathrm{H}_2\mathrm{S}\).
03

Determine stronger intermolecular forces based on boiling and melting points

Generally, higher boiling and melting points are an indication of stronger intermolecular forces in a substance. As the boiling and melting points of \(\mathrm{H}_2\mathrm{O}\) are higher than those of \(\mathrm{H}_2\mathrm{S}\), we can conclude that the intermolecular forces in \(\mathrm{H}_2\mathrm{O}\) are stronger.
04

Identify the types of intermolecular forces present in each molecule

The three main types of intermolecular forces are: London dispersion forces, dipole-dipole interactions, and hydrogen bonding. \(\mathrm{H}_2\mathrm{O}\) is a polar molecule due to the difference in electronegativities between Hydrogen and Oxygen atoms. As a result, \(\mathrm{H}_2\mathrm{O}\) has dipole-dipole interactions. In addition, Oxygen, being a highly electronegative atom, forms a strong hydrogen bond with a Hydrogen from another \(\mathrm{H}_2\mathrm{O}\) molecule. \(\mathrm{H}_2\mathrm{S}\) is also a polar molecule, meaning it has dipole-dipole interactions. However, while sulfur is more electronegative than hydrogen, sulfur's electronegativity is significantly lower than oxygen's, thus, forming weaker dipole-dipole interactions. Additionally, sulfur does not form hydrogen bonding like oxygen. In conclusion, it is the hydrogen bonding in \(\mathrm{H}_2\mathrm{O}\) that makes its intermolecular forces stronger compared to the dipole-dipole interactions in \(\mathrm{H}_2\mathrm{S}\).

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

A watch with a liquid crystal display (LCD) does not function properly when it is exposed to low temperatures during a trip to Antarctica. Explain why the LCD might not function well at low temperature.

True or false: (a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. (b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, dipole- dipole interactions, when present, are always greater than dispersion forces.( \(\mathbf{d}\) ) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The larger the atom, the more polarizable it is.

Describe the intermolecular forces that must be overcome to convert these substances from a liquid to a gas: (a) SO \(_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{COOH},(\mathbf{c}) \mathrm{H}_{2} \mathrm{S}\) .

Name the phase transition in each of the following situations and indicate whether it is exothermic or endothermic: (a) Bromine vapor turns to bromine liquid as it is cooled. (b) Crystals of iodine disappear from an evaporating dish as they stand in a fume hood. (c) Rubbing alcohol in an open container slowly disappears. (d) Molten lava from a volcano turns into solid rock.

In terms of the arrangement and freedom of motion of the molecules, how are the nematic liquid crystalline phase and an ordinary liquid phase similar? How are they different?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

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