Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 8, Problem 65E
Interpretation Introduction
Interpretation: The energy of sublimation is to be calculated from the given data of lithium iodide reaction.
Concept introduction: The sublimation is the process in which change of phase occur from solid to gas. The energy change occur during this process is called sublimation energy
To determine: The value of energy of sublimation for the given reaction.
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Calculate 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/mol
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, 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/mol
Chapter 8 Solutions
Chemistry
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- Calculate the second ionization energy of the metal M (AH¡on2 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 %3D 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), AHf = -341 kJ/molarrow_forwardConsider the following reaction:A2 + B2 → 2AB ΔH = –377 kJThe bond energy for A2 is half the amount of AB. The bond energy of B2 = 405 kJ/mol. What is the bond energy of A2?arrow_forwardUse the following data to estimate AH; for sodium bromide. Na(s) + Br2 (9) → NaBr(s) Lattice energy -742 kJ/mol Ionization energy for Na 495 kJ/mol -325 kJ/mol Electron affinity of Br Bond energy of Br2 Enthalpy of sublimation for Na 193 kJ/mol 109 kJ/mol AH; = | kJ/molarrow_forward
- 4. Referring to a periodic table, arrange the following atoms in order of increasing atomic radius, ionization energy and electronegativity: P, Si, N?arrow_forwardThe electron affinity of oxygen is -141kj/mol, corresponding to the reaction O(g)+e-—>O-(g) The lattice energy of K2O(s) is 2238kj/mol.Use these data along with data in Appendix C and figure 7.10 to calculate the “second electron affinity” of oxygen, corresponding to the reaction O-(g)+e-—>O2-(g)arrow_forwardConsider the following data for ruthenium: atomic mass 101.07 g/mol electronegativity 2.20 electron affinity 101.3 kJ/mol ionization energy 710.2 kJ/mol heat of fusion 25.7 kJ/mol Does the following reaction absorb or release energy? (1) Ru+(g) +e− -> Ru(g) Is it possible to calculate the amount of energy absorbedor released by reaction (1) using only the data above? If you answered yes to the previous question, enter theamount of energy absorbed or released by reaction (1): Does the following reaction absorb or release energy? (2) Ru−(g) -> Ru(g) + e− Is it possible to calculate the amount of energy absorbedor released by reaction (2) using only the data above? If you answered yes to the previous question, enter theamount of energy absorbed or released by reaction (2)arrow_forward
- Suppose there is an element X which occurs naturally as X2(g).X2(g) + 2O2(g) → X2O4(g)ΔHof of O(g) is 249 kJ/molΔHof of X(g) is 458.5 kJ/molΔHof of X2O4(g) is 31 kJ/molThe X-X single bond energy is 116 kJ/molUse the above data to estimate the average bond energy in X2O4. Give your answer to the nearest 1 kJ/mol.arrow_forwardThe following Lewis diagram represents the valence electron configuration of a main-group element. This element is in group [VIIA(17) According to the octet rule, this element would be expected to form an ion with a charge of 1- If X is in period 4, the ion formed has the same electron configuration as the noble gas The symbol for the ion isarrow_forward2. Calculate the lattice energy of MgO, given the following: Mg(s) + ¼O:(g) → Mg0(s) AH = -602 kJ AH = 150 kJ AH = 737 kJ Mg(s) → Mg(g) O(g) + 2e (g) → 0*(g) 20(g)→0:(g) Mg(g) → Mg*(g) + 2 e (g) AH = -494 kJ AH = 2180 kJarrow_forward
- Write 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_forwardCalculate the lattice enthalpy (DeltaH in kJ/mol) of the MF2 metal fluoride using the following data: Bond dissociation enthalpy of F2(g)=159 kJ/mol Electron affinity of F = -328 kJ/mol Enthalpy of sublimation of M = 108 kJ/mol First ionization energy of M = 412 kJ/mol Second ionization energy of M = 716 kJ/mol Standard enthalpy of formation of MF2(s) = -376 kJ/molarrow_forwardThe first four ionization energies of an element X are 578, 1817, 2745, and 11,577 kJ·mol–1. What is the most likely formula for the most stable ion of Xarrow_forward
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