Ch.8 - Basic Concepts of Chemical Bonding

Brown - Chemistry: The Central Science 15th Edition

Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Not the one you use?Change textbook

Brown 15th Edition

Ch.8 - Basic Concepts of Chemical Bonding

Problem 31

Chapter 8, Problem 31

Use data from Appendix C, Figure 7.11, and Figure 7.13 to calculate the lattice energy of KI.

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Identify the Born-Haber cycle for KI, which involves the following steps: sublimation of K(s) to K(g), ionization of K(g) to K^+(g), dissociation of I2(g) to 2 I(g), electron affinity of I(g) to I^-(g), and formation of KI(s) from K^+(g) and I^-(g).

Use Appendix C to find the enthalpy of sublimation for K(s) to K(g).

Use Appendix C to find the ionization energy for K(g) to K^+(g).

Use Appendix C to find the bond dissociation energy for I2(g) to 2 I(g), and then calculate the energy for I(g) to I^-(g) using the electron affinity.

Apply Hess's Law to sum the enthalpies of each step in the Born-Haber cycle, including the lattice energy, to solve for the lattice energy of KI.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lattice Energy

Lattice energy is the amount of energy released when gaseous ions combine to form an ionic solid. It is a measure of the strength of the forces between the ions in an ionic compound. Higher lattice energy indicates stronger ionic bonds, which typically results in higher melting points and greater stability of the compound.

Born-Haber Cycle

The Born-Haber cycle is a thermodynamic cycle that relates the lattice energy of an ionic compound to other energy changes involved in its formation. It includes steps such as ionization energy, electron affinity, and sublimation energy, allowing for the calculation of lattice energy using Hess's law. This cycle is essential for understanding how different energy contributions affect the overall stability of ionic compounds.

Ionic Radii

Ionic radii refer to the effective size of an ion in a crystal lattice. The size of the ions affects the distance between them, which in turn influences the lattice energy. Smaller ions can pack more closely together, leading to stronger electrostatic attractions and higher lattice energies, while larger ions result in weaker attractions and lower lattice energies.

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Related Practice

Textbook Question

Energy is required to remove two electrons from Ca to form Ca2+, and energy is required to add two electrons to O to form O2 - . Yet CaO is stable relative to the free elements. Which statement is the best explanation? (a) The lattice energy of CaO is large enough to overcome these processes. (b) CaO is a covalent compound, and these processes are irrelevant. (c) CaO has a higher molar mass than either Ca or O. (d) The enthalpy of formation of CaO is small. (e) CaO is stable to atmospheric conditions.

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Textbook Question

Which of the following trends in lattice energy is due to differences in ionic radii: a. NaCl > RbBr > CsBr, b. BaO > KF, c. SrO > SrCl2?

Textbook Question

List the individual steps used in constructing a Born–Haber cycle for the formation of BaI₂ from the elements. Which of the steps would you expect to be exothermic?

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Textbook Question

(a) Based on the lattice energies of MgCl2 and SrCl2 given in Table 8.1, what is the range of values that you would expect for the lattice energy of CaCl2?

Textbook Question

(b) Using data from Appendix C, Figure 7.11, Figure 7.13, and the value of the second ionization energy for Ca, 1145 kJ/mol, calculate the lattice energy of CaCl2.

Textbook Question

(b) A substance, XY, formed from two different elements, melts at −33 °C. Is XY likely to be a covalent or an ionic substance?

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