Student Solutions Manual for Ball's Physical Chemistry, 2nd
2nd Edition
ISBN: 9798214169019
Author: David W. Ball
Publisher: Cengage Learning US
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Chapter 21, Problem 21.53E
Interpretation Introduction
Interpretation:
The lattice energy of
Concept introduction:
The amount of energy released when one formula unit moles of oppositely charged gaseous ions binds together to form a crystal is known as the Lattice energy. The value of lattice energy is negative. It is used as the measure for stability of a crystal.
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Students have asked these similar questions
Calculate the lattice energy of TIF (s) using the
following thermodynamic data (all data is in
kJ/mol). Note that the data given has been
perturbed, so looking up the answer is
probably not a good idea.
TI (S)
TI (g)
F - F
(g)
F (g)
TIF
(s)
AHsublimation = 161
kJ/mol
Ionization energy =
569 kJ/mol
Bond energy = 138
kJ/mol
Electron affinity = -348
kJ/mol
AH°f = -345 kJ/mol
kJ/mol
The ionisation energy of potassium is 4.34 eV and the electron affinity of chlorine is 3.61 eV.
The Madelung constant for the KCl structure is 1.748 and the closest distance between ions of opposite sign is 0.314 nm.
On the basis of these data, calculate the cohesive energy of KCl. Compare this with the observed cohesive energy of 6.42 eV for the ion pair and comment on the reasons for any discrepancy
Given the following thermodynamic data, calculate the lattice energy of LiCl:ΔH°f[LiCl(s)] = -409 kJ/molΔH°sublimation [Li] = 161 kJ/molBond energy [Cl-Cl] = 243 kJ/molIE1 (Li) = 520 kJ/molEA1 (Cl) = -349 kJ/mol
-1682 kJ/mol
-984 kJ/mol
-1560 kJ/mol
-862 kJ/mol
-1213 kJ/mol
Chapter 21 Solutions
Student Solutions Manual for Ball's Physical Chemistry, 2nd
Ch. 21 - Prob. 21.1ECh. 21 - Boron nitride, BN, is a very hard material, harder...Ch. 21 - Prob. 21.3ECh. 21 - Prob. 21.4ECh. 21 - Figure 21.35 shows a unit cell of diamond....Ch. 21 - Prob. 21.6ECh. 21 - How many different unit cells can a crystal have...Ch. 21 - Prob. 21.8ECh. 21 - Prob. 21.9ECh. 21 - Prob. 21.10E
Ch. 21 - Prob. 21.11ECh. 21 - Prob. 21.12ECh. 21 - Prob. 21.13ECh. 21 - Prob. 21.14ECh. 21 - Prob. 21.15ECh. 21 - Prob. 21.16ECh. 21 - Prob. 21.17ECh. 21 - Prob. 21.18ECh. 21 - Prob. 21.19ECh. 21 - Prob. 21.20ECh. 21 - Prob. 21.21ECh. 21 - Prob. 21.22ECh. 21 - Prob. 21.23ECh. 21 - Prob. 21.24ECh. 21 - Prob. 21.25ECh. 21 - Prob. 21.26ECh. 21 - Prob. 21.27ECh. 21 - Prob. 21.28ECh. 21 - For a simple cubic lattice, what is the ratio of...Ch. 21 - Prob. 21.30ECh. 21 - Prob. 21.31ECh. 21 - Consider Figure 21.21. If the lower rightmost...Ch. 21 - Prob. 21.33ECh. 21 - The aluminum-nickel alloy AlNi has a simple cubic...Ch. 21 - Prob. 21.35ECh. 21 - The first two signals from a powdered sample has X...Ch. 21 - Prob. 21.37ECh. 21 - Prob. 21.38ECh. 21 - Prob. 21.39ECh. 21 - Prob. 21.40ECh. 21 - Prob. 21.41ECh. 21 - Prob. 21.42ECh. 21 - Prob. 21.43ECh. 21 - Prob. 21.44ECh. 21 - Prob. 21.45ECh. 21 - What is the coordination number in the cesium...Ch. 21 - Prob. 21.47ECh. 21 - Which solid phase that is, which allotrope of...Ch. 21 - Prob. 21.49ECh. 21 - Prob. 21.50ECh. 21 - Write Born-Haber cycles showing the relationship...Ch. 21 - Prob. 21.52ECh. 21 - Prob. 21.53ECh. 21 - Prob. 21.54ECh. 21 - The lattice energy for potassium iodide, KI, is...Ch. 21 - Prob. 21.56ECh. 21 - Prob. 21.57ECh. 21 - Prob. 21.58ECh. 21 - Prob. 21.59ECh. 21 - Prob. 21.60ECh. 21 - Prob. 21.61ECh. 21 - Prob. 21.62ECh. 21 - Prob. 21.63ECh. 21 - Prob. 21.64E
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- Calculate the standard molar enthalpy of formation of NO(g) from the following data: N2(g)+2O22NO2(g)H298=66.4kJ2NO(g)+O22NO2(g)H298=114.1kJarrow_forwardExplain the term crystal radii?arrow_forward16). Using the thermochemical data below calculate the lattice energy for the formation of Na₂O. Na(s) → Na(g) 107.3 kJ/mol Na(g) → Na (g) + 1 e - 495.9 kJ/mol -418 kJ/mol 249.1 kJ/mol -141 kJ/mol -1484.5 kJ/mol 2 Na(s) + O₂(g) →→→ Na₂O(s) 1/2 O₂(g) → 0(g) O(g) + 1 e→O(g) O(g) +1e0²(g)arrow_forward
- In a hydrogen molecule, the two hydrogen atoms are held together by a single bond with a bond energy of 436 kJ/mol of hydrogen. In other words, to break the H-H bonds in one mole of molecular hydrogen requires the expenditure of 436 kJ of energy. Using the balanced chemical equation for the formation of water from oxygen and hydrogen (shown above), and interpreting the stoichiometric coefficients as mole amounts, how much energy must be expended in breaking the H-H bonds? kJarrow_forwardCalculate the lattice enthalpy for RbC1. You will need the following information: Species AfH°, kJ/mol Rb(g) RbCl(s) Cl(g) 80.9 - 435.4 121.3 Enthalpy of ionization for Rb(g) is 403.0 kJ/mol; electron attachment enthalpy for Cl(g) is −349.0 kJ/mol. Lattice enthalpy = kJ/molarrow_forwardIn a lattice, a positive ion is often surrounded by eight negative ions. We might reason, therefore, that the lattice energy should be related to eight times the potential of interaction between these oppositely charged particles. Why is this reasoning too simple?arrow_forward
- Calculate the lattice energy of RbH (s) using the following thermodynamic data (all data is in kJ/mol). Note that the data given has been perturbed, so looking up the answer is probably not a good idea. AHsublimation 61 kJ/mol Ionization energy = 383 kJ/mol Rb (s) Rb (g) H - H (g) Bond energy = 416 kJ/mol H (g) Electron affinity = -93 kJ/mol RbH (s) AHOf = -72 kJ/mol kJ/molarrow_forwardTo explain why AHan for RbCl is different than that for KCI, the student investigates factors that affect AHoln and finds that ionic radius and lattice enthalpy (which can be defined as the AH associated with the separation of a solid crystal into gaseous ions) contribute to the process. The students consults references and collects the data shown in the table below. lonic Radius (pm) lon к' 138 Rb* 152 b. Using principles of atomic structure, explain why the Rb' ion is larger than the K' ionarrow_forwardCalculate the lattice energy of AgCl(s) using the following thermodynamic data (all data is in kJ/mol). Note that the data given has been perturbed, so looking up the answer is probably not a good idea. Ag(s) AHsublimation 265 kJ/mol Ag(g) Ionization energy 711 kJ/mol Cl-CI(g) Bond energy = 223 kJ/mol CI(g) Electron affinity = -369 kJ/mol AgCl(s) AH°=-147 kJ/mol kJ/mol Do you expect this value to be larger or smaller than the lattice energy of AgBr(s)?arrow_forward
- Calculate the lattice energy,U (in kJ), of the fictitious compound, MCl2, given the following information: ΔHppt for MCl2 = 17.1 kJ/mol, ΔHhyd for M2+ = -450 kJ/mol, ΔHhyd for Cl- = -370 kJ/mol Your answer should have 4 sig figs.arrow_forwardCalculate the lattice energy of AgBr(s) using the following thermodynamic data (all data is in kJ/mol). Ag(s) ΔHsublimation = 265 kJ/mol Ag(g) Ionization energy = 711 kJ/mol Br-Br(g) Bond energy = 173 kJ/mol Br(g) Electron affinity = -345 kJ/mol AgBr(s) ΔH°f = -120 kJ/mol ______________ kJ/molarrow_forwardlattice energy 1.0arrow_forward
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