EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 9781305856745
Author: DECOSTE
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 13, Problem 41E
Consider the following:
of sublimation of Li(s).
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Use the Born-Haber cycle to calculate the lattice energy of KF. [The heat of sublimation of K is 91.6 kJ·mol−1 and
ΔfH(KF) = −567.3 kJ·mol−1.
Bond enthalpy for F2 is
158.8 kJ·mol−1.
Other data may be found in the Ionization Energies Table and the Electron Affinities Table.]
Using the following data, calculate the lattice energy of calcium chloride:
Ca2+(g) + 2Cl– (g) → CaCl2(s) ΔHlattice = ?
Sublimation enthalpy of calcium ΔH = 177.8 kJ/mol
First ionization energy of calcium ΔH = 590.2 kJ/mol
Second ionization energy of calcium ΔH = 1144.2 kJ/mol
First electron affinity of chlorine ΔH = –349 kJ/mol
Heat of formation of CaCl2(s) ΔH = –795.4 kJ/mol
Bond energy of Cl2 (see Table 2)
Use Hess’s law to calculate the lattice energy of calcium chloride. set-up must show all the chemical equations and you must show how their H values add up to give your answer.
Calculate the lattice energy for LiBr(s) given the following:
sublimation energy for Li(s)
+166 kJ/mol
ΔHf for Br(g)
+97 kJ/mol
first ionization energy of Li(g)
+520. kJ/mol
electron affinity of Br(g)
–325 kJ/mol
enthalpy of formation of LiBr(s)
–351 kJ/mol
Chapter 13 Solutions
EBK CHEMICAL PRINCIPLES
Ch. 13 - Explain the electronegativity trends across a row...Ch. 13 - Prob. 2DQCh. 13 - Prob. 3DQCh. 13 - Prob. 4DQCh. 13 - Prob. 5DQCh. 13 - Prob. 6DQCh. 13 - Prob. 7DQCh. 13 - Prob. 8DQCh. 13 - Prob. 9DQCh. 13 - Arrange the following molecules from most to least...
Ch. 13 - Prob. 11DQCh. 13 - Prob. 12DQCh. 13 - Prob. 13ECh. 13 - Prob. 14ECh. 13 - An alternative definition of electronegativity...Ch. 13 - Prob. 16ECh. 13 - Without using Fig. 13.3, predict the order of...Ch. 13 - Without using Fig. 13.3, predict which bond in...Ch. 13 - Prob. 19ECh. 13 - Prob. 20ECh. 13 - Indicate the bond polarity (show the partial...Ch. 13 - Prob. 22ECh. 13 - Prob. 23ECh. 13 - Prob. 24ECh. 13 - Prob. 25ECh. 13 - Prob. 26ECh. 13 - Prob. 27ECh. 13 - Prob. 28ECh. 13 - Prob. 29ECh. 13 - Prob. 30ECh. 13 - Prob. 31ECh. 13 - Give an example of an ionic compound where both...Ch. 13 - What noble gas has the same electron configuration...Ch. 13 - Which of the following ions have noble gas...Ch. 13 - Give three ions that are isoelectronic with...Ch. 13 - Prob. 36ECh. 13 - Predict the empirical formulas of the ionic...Ch. 13 - Which compound in each of the following pairs of...Ch. 13 - Use the following data to estimate Hf for...Ch. 13 - Use the following data to estimate Hf for...Ch. 13 - Consider the following:...Ch. 13 - In general, the higher the charge on the ions in...Ch. 13 - Consider the following energy changes: a....Ch. 13 - Prob. 44ECh. 13 - Prob. 45ECh. 13 - The lattice energies of FeCl3,FeCl2,andFe2O3 are...Ch. 13 - Prob. 47ECh. 13 - Prob. 48ECh. 13 - Prob. 49ECh. 13 - Prob. 50ECh. 13 - Prob. 51ECh. 13 - Prob. 52ECh. 13 - Prob. 53ECh. 13 - Prob. 54ECh. 13 - Prob. 55ECh. 13 - Prob. 56ECh. 13 - Prob. 57ECh. 13 - Prob. 58ECh. 13 - Prob. 59ECh. 13 - Prob. 60ECh. 13 - Prob. 61ECh. 13 - Prob. 62ECh. 13 - Prob. 63ECh. 13 - Prob. 64ECh. 13 - Prob. 65ECh. 13 - Prob. 66ECh. 13 - Prob. 67ECh. 13 - Prob. 68ECh. 13 - Prob. 69ECh. 13 - Prob. 70ECh. 13 - Prob. 71ECh. 13 - Prob. 72ECh. 13 - Prob. 73ECh. 13 - Prob. 74ECh. 13 - Prob. 75ECh. 13 - Prob. 76ECh. 13 - Prob. 77ECh. 13 - Prob. 78ECh. 13 - Prob. 79ECh. 13 - Prob. 80ECh. 13 - Prob. 81ECh. 13 - Prob. 82ECh. 13 - Prob. 83ECh. 13 - Prob. 84ECh. 13 - Prob. 85ECh. 13 - Prob. 86ECh. 13 - Prob. 87ECh. 13 - Prob. 88ECh. 13 - Prob. 89ECh. 13 - Prob. 90ECh. 13 - Prob. 91ECh. 13 - Prob. 92ECh. 13 - Prob. 93ECh. 13 - Prob. 94ECh. 13 - Prob. 95ECh. 13 - Predict the molecular structure and the bond...Ch. 13 - Prob. 97ECh. 13 - Two variations of the octahedral geometry are...Ch. 13 - Prob. 99ECh. 13 - Predict the molecular structure and the bond...Ch. 13 - Which of the molecules in Exercise 96 have net...Ch. 13 - Prob. 102ECh. 13 - Give two requirements that should be satisfied for...Ch. 13 - What do each of the following sets of...Ch. 13 - Prob. 105ECh. 13 - Consider the following Lewis structure, where E is...Ch. 13 - Consider the following Lewis structure, where E is...Ch. 13 - Prob. 108ECh. 13 - Prob. 109ECh. 13 - Which of the following molecules have net dipole...Ch. 13 - Prob. 111AECh. 13 - Prob. 112AECh. 13 - Prob. 113AECh. 13 - Prob. 114AECh. 13 - Prob. 115AECh. 13 - There are two possible structures of XeF2Cl2 ,...Ch. 13 - Prob. 117AECh. 13 - Prob. 118AECh. 13 - Prob. 119AECh. 13 - Prob. 120AECh. 13 - Prob. 121AECh. 13 - Prob. 122AECh. 13 - Prob. 123AECh. 13 - Prob. 124AECh. 13 - Prob. 125AECh. 13 - Prob. 126AECh. 13 - Prob. 127AECh. 13 - Prob. 128AECh. 13 - Prob. 129AECh. 13 - Prob. 130AECh. 13 - Prob. 131AECh. 13 - Prob. 132AECh. 13 - Prob. 133CPCh. 13 - Prob. 134CPCh. 13 - Given the following information: Heat of...Ch. 13 - Prob. 136CPCh. 13 - A promising new material with great potential as...Ch. 13 - Think of forming an ionic compound as three steps...Ch. 13 - Prob. 139CPCh. 13 - Prob. 140CPCh. 13 - Calculate the standard heat of formation of the...Ch. 13 - Prob. 142CPCh. 13 - Prob. 143MP
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- Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in CS2.arrow_forwardCalculate the lattice energy of potassium fluoride, KF, using the BornHaber cycle. Use thermodynamic data from Appendix C to obtain the enthalpy changes for each step. (Note: You will obtain a slightly different answer if you use values given in Chapter 8 for the ionization energy and electron affinity, which are energy values at 0 K rather than the enthalpy changes at 298 K.)arrow_forwardBond Enthalpy When atoms of the hypothetical element X are placed together, they rapidly undergo reaction to form the X2 molecule: X(g)+X(g)X2(g) a Would you predict that this reaction is exothermic or endothermic? Explain. b Is the bond enthalpy of X2 a positive or a negative quantity? Why? c Suppose H for the reaction is 500 kJ/mol. Estimate the bond enthalpy of the X2 molecule. d Another hypothetical molecular compound, Y2(g), has a bond enthalpy of 750 kJ/mol, and the molecular compound XY(g) has a bond enthalpy of 1500 kJ/mol. Using bond enthalpy information, calculate H for the following reaction. X2(g)+Y2(g)2XY(g) e Given the following information, as well as the information previously presented, predict whether or not the hypothetical ionic compound AX is likely to form. In this compound, A forms the A+ cation, and X forms the X anion. Be sure to justify your answer. Reaction: A(g)+12X2(g)AX(s)The first ionization energy of A(g) is 400 kJ/mol. The electron affinity of X(g) is 525 kJ/mol. The lattice energy of AX(s) is 100 kJ/mol. f If you predicted that no ionic compound would form from the reaction in Part e, what minimum amount of AX(s) lattice energy might lead to compound formation?arrow_forward
- 8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of thesearrow_forwardUsing the following data, estimate the overall enthalpy of formation (in kJ/mol) for potassium chloride: K(s) + ½ Cl₂(g) → KCI(s). Process Lattice energy of KCI lonization energy of K Electron affinity of Cl Bond dissociation energy of Cl, Enthalpy of sublimation for K Question 21 of 28 Change in Energy (AHO) -690 kJ/mol 419 kJ/mol -349 kJ/mol 239 kJ/mol 90 kJ/molarrow_forwardGiven the following information: Heat of sublimation of Li(s) = 166 kJ/mol Bond energy of HF = 565 kJ/mol Ionization energy of Li(g) = 520. kJ/mol Electron affinity of F(g) = –328 kJ/mol Lattice energy of LiF(s) = –1030 kJ/mol Bond energy of H2 = 432 kJ/mol Calculate the net change in energy for the following reaction: 2Li(s)+2HF(g)-> 2LiF(s) +H2(g) Change in energy = kJarrow_forward
- 5. Consider the following information: 1st ionization energy of Na(g) = 495.8 kJ/mol Bond dissociation energy of O2(g) = 498.4 kJ/mol 1st electron affinity of O(g)=-142.5 kJ/mol 2nd electron affinity of O¹(g) = 844 kJ/mol Lattice energy of Na2O(s) = -2608 kJ/mol Enthalpy of formation of Na2O(s) = -416 kJ/mol a Draw the Born-Haber cycle for Na₂O(s). b Calculate the unknown. 120 C Draw the Lewis symbol for Na₂O.arrow_forwardCalculate the second ionization energy of the metal M (AHion2 in kJ/mol) using the following data: Lattice enthalpy of MO(s), AH = -2278 kJ/mol Bond dissociation enthalpy of O2(g) = +498 kJ/mol First electron affinity of O = -141 kJ/mol Second electron affinity of O = +744 kJ/mol Enthalpy of sublimation of M = + 125 kJ/mol First ionization energy of M = + 309 kJ/mol Standard enthalpy of formation of MO(s), AH = -341 kJ/molarrow_forwardCalculate the lattice energy of NaBr(s), given the following thermochemical equations, where A/E and AEA are ionization energy and electron affinity, respectively. Na(s)Na(g) AH = +107 kJ Na(g) Nat(g) + e A/E = +496 kJ -> 1/2 Br₂(g) → Br(g) AHf = +112 kJ - Br(g) + e¯ → Br¯(g) AEA = -325 kJ Na(s) + 1/2 Br₂(g) → NaBr(s) AH = -361 kJ ->> - -1401 kJ -751 kJ +29 kJ -29 kJ +751 kJarrow_forward
- The standard heat of formation of BaBr2BaBr2 is −−764 kJ/molkJ/mol. The first ionization energy of BaBa is 503 kJ/molkJ/mol and its second ionization energy is 965 kJ/molkJ/mol. The heat of sublimation of Ba[Ba(s)→Ba(g)]Ba[Ba(s)→Ba(g)] is 175 kJ/molkJ/mol. The bond energy of Br2Br2 is 193 kJ/molkJ/mol, the heat of vaporization of Br2(l)Br2(l) is 31 kJ/molkJ/mol, and the electron affinity of BrBr is −−325 kJ/molkJ/mol. Calculate the lattice energy of BaBr2BaBr2.arrow_forwardChoose the related energy for the following reaction: 2 Cs* (g) + O2- (g) → Cs20 (s) electron affinity ionization energy heat of formation lattice energyarrow_forwardWrite the steps (reactions) for the Born-Haber cycle for MgCl2(s). Use the Born-Haber cycle to calculate the lattice energy of MgCl2(s). Some useful data to work with: For Mg: ΔΔHsub = 147 kJ/mol, IE1 and IE2 are 738 kJ/mol and 1450 kJ/mol, respectively. For chlorine: Bond energy = 243 kJ/mol, EA1 = -349 kJ/mol, respectively. The enthalpy of formation of magnesium chloride is -748.8 kJ/mol.arrow_forward
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