/*! 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 86 9.86 Lattice energies can also b... [FREE SOLUTION] | 91影视

91影视

9.86 Lattice energies can also be calculated for covalent network solids using a Born-Haber cycle, and the network solid silicon dioxide has one of the highest \(\Delta H_{\text {latice }}^{\circ}\) values. Silicon dioxide is found in pure crystalline form as transparent rock quartz. Much harder than glass, this material was once prized for making lenses for optical devices and expensive spectacles. Use Appendix \(\mathrm{B}\) and the following data to calculate \(\Delta H_{\text {lattioe }}^{\circ}\) of \(\mathrm{SiO}_{2}\) $$ \begin{array}{ll} \operatorname{Si}(s) \longrightarrow \operatorname{Si}(g) & \Delta H^{\circ}=454 \mathrm{~kJ} \\ \mathrm{Si}(g) \longrightarrow \mathrm{Si}^{4+}(g)+4 \mathrm{e}^{-} & \Delta H^{\circ}=9949 \mathrm{~kJ} \\ \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) & \Delta H^{\circ}=498 \mathrm{~kJ} \\ \mathrm{O}(g)+2 \mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(g) & \Delta H^{\circ}=737 \mathrm{~kJ} \end{array} $$

Short Answer

Expert verified
12375 kJ

Step by step solution

01

Sublimation of Silicon

Convert solid silicon to gaseous silicon using the given data \ \[ \text{Si}(s) \rightarrow \text{Si}(g) \ \text{螖H} = 454 \text{ kJ} \]
02

Ionization of Silicon

Convert gaseous silicon to silicon ion by removing four electrons \ \[ \text{Si}(g) \rightarrow \text{Si}^{4+}(g) + 4e^{-} \ \text{螖H} = 9949 \text{ kJ} \]
03

Dissociation of Oxygen

Dissociate molecular oxygen into two oxygen atoms \ \[ \text{O}_2(g) \rightarrow 2\text{O}(g) \ \text{螖H} = 498 \text{ kJ} \]
04

Formation of Oxide Ions

Convert oxygen atoms to oxide ions by adding electrons \ \[ \text{O}(g) + 2e^{-} \rightarrow \text{O}^{2-}(g) \ \text{螖H} = 737 \text{ kJ} \ \text{For 2 atoms, 螖H = 2 \times 737 = 1474 kJ} \]
05

Combine the Steps

Add the enthalpy changes from steps 1 to 4 to find the total enthalpy change associated with forming one mole of silicon dioxide ions \ \[ \text{螖H}_{\text{total}} = 454 + 9949 + 498 + 1474 = 12375 \text{ kJ} \]

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 91影视!

Key Concepts

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

Born-Haber Cycle
The Born-Haber Cycle is a concept used in thermochemistry to analyze and calculate the lattice energy of ionic compounds. It uses a series of hypothetical steps and Hess's Law to derive the energy changes that take place during the formation of an ionic solid from its gaseous ions. Each step in the cycle represents a physical or chemical change with a known enthalpy change. The key steps typically involve ionization, electron affinity, sublimation, and bond formation. Understanding the Born-Haber Cycle allows us to quantify the energy associated with the formation of solid ionic compounds and predict the stability of the formed crystals.
Covalent Network Solids
Covalent network solids are materials where atoms are bonded by covalent bonds in a continuous network that extends throughout the material. Unlike molecular solids where molecules are held together by weaker forces, covalent network solids have high melting points and are very hard due to the strength and directionality of covalent bonds. Diamond and silicon dioxide (SiO鈧) are classic examples of covalent network solids. The extended bonding network results in unique properties like hardness and thermal stability. This type of solid is incredibly stable, making them ideal for industrial applications where durability and resistance to high temperatures are needed.
Silicon Dioxide Thermodynamics
Silicon dioxide (SiO鈧), commonly found in nature as quartz, has significant thermodynamic properties that influence its stability and usage. The high lattice energy of SiO鈧 reflects its strong covalent bonds arranged in a 3D network. In its crystalline form, SiO鈧 is highly ordered with each silicon atom covalently bonded to four oxygen atoms. The thermodynamics involve several stages: sublimation of silicon, ionization of silicon, dissociation of oxygen molecules, and formation of oxide ions. These steps combined give the substance its high thermodynamic stability, making it resistant to melting and breaking apart at standard conditions.
Ionization Energy
Ionization energy is the energy required to remove an electron from a gaseous atom or ion. It is a crucial concept in understanding the reactivity and formation of ions. In the context of lattice energy calculations, the ionization energy of silicon is significant because transforming Si to Si鈦粹伜 involves a substantial amount of energy. Multiple ionization steps illustrate increasing energy requirements as electrons are progressively removed, which collectively contribute to the total enthalpy change in the Born-Haber Cycle. A higher ionization energy generally indicates a more stable atom to its electrons, affecting the overall process of forming ionic compounds.
Sublimation
Sublimation refers to the phase transition of a substance from a solid directly to a gas without passing through the liquid phase. It is an endothermic process requiring energy input to overcome intermolecular forces. In the context of the exercise, sublimation of solid silicon (Si) is the first step, where silicon transitions to its gaseous form. This step is essential as it sets the stage for further ionization processes in the Born-Haber Cycle. Specifically, the enthalpy change for sublimation of silicon provides the energy needed to break Si-Si bonds and convert solid silicon into single, gaseous silicon atoms.
Dissociation Energy
Dissociation energy is the energy required to break a specific bond in a molecule leading to the formation of smaller species, usually atoms or radicals. It's a measure of bond strength in a chemical bond. For molecular oxygen (O鈧), the dissociation energy involves breaking the O=O double bond to yield two oxygen atoms. This process is a critical step in the Born-Haber Cycle for calculating the lattice energy of SiO鈧, as it provides the necessary oxygen atoms that will later combine with silicon ions. High dissociation energy values reflect stronger bonds, highlighting the considerable energy input needed to disrupt such stable molecules.

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

Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent? Arrange the substances with polar covalent bonds in order of increasing bond polarity: (a) KCl (b) \(\mathrm{P}_{4}\) (c) BF \(_{3}\) (d) \(\mathrm{SO}_{2}\) (e) \(\mathrm{Br}_{2}\) (f) \(\mathrm{NO}_{2}\)

Use the following to calculate \(\Delta H_{\text {latice }}^{\circ}\) of \(\mathrm{MgF}_{2}\) : \(\begin{array}{llr}\mathrm{Mg}(s) \longrightarrow \mathrm{Mg}(g) & \Delta H^{\circ}= & 148 \mathrm{~kJ} \\ \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g) & \Delta H^{\circ}= & 159 \mathrm{~kJ} \\ \mathrm{Mg}(g) \longrightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-} & \Delta H^{\circ}= & 738 \mathrm{~kJ} \\ \mathrm{Mg}^{+}(g) \longrightarrow \mathrm{Mg}^{2+}(g)+\mathrm{e}^{-} & \Delta H^{\circ}= & 1450 \mathrm{~kJ} \\\ \mathrm{~F}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{F}^{-}(g) & \Delta H^{\circ}= & -328 \mathrm{~kJ} \\ \mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s) & \Delta H^{\circ}=-1123 \mathrm{~kJ}\end{array}\) Compared with the lattice energy of LiF ( \(1050 \mathrm{~kJ} / \mathrm{mol}\) ) or the lattice energy you calculated for \(\mathrm{NaCl}\) in Problem \(9.30,\) does the relative magnitude of the value for \(\mathrm{MgF}_{2}\) surprise you? Explain.

Using the periodic table only, arrange the members of each of the following sets in order of increasing bond length: (a) \(\mathrm{H}-\mathrm{F}, \mathrm{H}-\mathrm{I}, \mathrm{H}-\mathrm{Cl}\) (b) \(\mathrm{C}-\mathrm{S}, \mathrm{C}=\mathrm{O}, \mathrm{C}-\mathrm{O}\) (c) \(\mathrm{N}-\mathrm{H}, \mathrm{N}-\mathrm{S}, \mathrm{N}-\mathrm{O}\)

The text points out that, for similar types of substances, one with weaker bonds is usually more reactive than one with stronger bonds. Why is this generally true?

Aluminum oxide \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) is a widely used industrial abrasive (known as emery or corundum), for which the specific application depends on the hardness of the crystal. What does this hardness imply about the magnitude of the lattice energy? Would you have predicted from the chemical formula that \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is hard? Explain.

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.