Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 130

For each of the following, write an equation that corresponds to the energy given. a. lattice energy of \(\mathrm{NaCl}\) b. lattice energy of \(\mathrm{NH}_{4} \mathrm{Br}\) c. lattice energy of \(\mathrm{MgS}\) d. \(\mathrm{O}=\mathrm{O}\) double bond energy beginning with \(\mathrm{O}_{2}(g)\) as a reactant

Short Answer

Expert verified
a. \[ \mathrm{NaCl(s)} \xrightarrow{lattice \: energy} \mathrm{Na^{+}(g)} + \mathrm{Cl^{-}(g)} \] b. \[ \mathrm{NH_{4}Br(s)} \xrightarrow{lattice \: energy} \mathrm{NH_{4}^{+}(g)} + \mathrm{Br^{-}(g)} \] c. \[ \mathrm{MgS(s)} \xrightarrow{lattice \: energy} \mathrm{Mg^{2+}(g)} + \mathrm{S^{2-}(g)} \] d. \[ \mathrm{O}_{2}(g) \xrightarrow{O=O \: double \: bond \: energy} 2\,\mathrm{O(g)} \]

Step by step solution

01

a. Lattice energy of NaCl

The lattice energy of \(\mathrm{NaCl}\) is the energy required to separate one mole of solid \(\mathrm{NaCl}\) into gaseous \(\mathrm{Na^{+}}\) ions and \(\mathrm{Cl^{-}}\) ions. We can write the equation as: \[ \mathrm{NaCl(s)} \xrightarrow{lattice \: energy} \mathrm{Na^{+}(g)} + \mathrm{Cl^{-}(g)} \]
02

b. Lattice energy of NH4Br

The lattice energy of \(\mathrm{NH_{4}Br}\) is the energy required to separate one mole of solid \(\mathrm{NH_{4}Br}\) into gaseous \(\mathrm{NH_{4}^{+}}\) ions and \(\mathrm{Br^{-}}\) ions. We can write the equation as: \[ \mathrm{NH_{4}Br(s)} \xrightarrow{lattice \: energy} \mathrm{NH_{4}^{+}(g)} + \mathrm{Br^{-}(g)} \]
03

c. Lattice energy of MgS

The lattice energy of \(\mathrm{MgS}\) is the energy required to separate one mole of solid \(\mathrm{MgS}\) into gaseous \(\mathrm{Mg^{2+}}\) ions and \(\mathrm{S^{2-}}\) ions. We can write the equation as: \[ \mathrm{MgS(s)} \xrightarrow{lattice \: energy} \mathrm{Mg^{2+}(g)} + \mathrm{S^{2-}(g)} \]
04

d. O=O double bond energy beginning with O2(g) as a reactant

The O=O double bond energy is the energy required to dissociate one mole of \(\mathrm{O}_{2}\) molecules with a double bond between its oxygen atoms. We can write the equation as: \[ \mathrm{O}_{2}(g) \xrightarrow{O=O \: double \: bond \: energy} 2\,\mathrm{O(g)} \]

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

What do each of the following sets of compounds/ions have in common with each other? a. \(\mathrm{SO}_{3}, \mathrm{NO}_{3}^{-}, \mathrm{CO}_{3}^{2-}\) b. \(\mathrm{O}_{3}, \mathrm{SO}_{2}, \mathrm{NO}_{2}^{-}\)

What is meant by a chemical bond? Why do atoms form bonds with each other? Why do some elements exist as molecules in nature instead of as free atoms?

Would you expect the electronegativity of titanium to be the same in the species \(\mathrm{Ti}, \mathrm{Ti}^{2+}, \mathrm{Ti}^{3+}\), and \(\mathrm{Ti}^{4+} ?\) Explain.

Write electron configurations for the most stable ion formed by each of the elements Rb, Ba, Se, and I (when in stable ionic compounds).

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}, \mathrm{O}-\mathrm{H}\), \(\mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Physical 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