General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 9, Problem 9.118SP
Interpretation Introduction

Interpretation:

Using the Born-Haber cycle for MgO lattice energy of O has to be calculated.

Concept Introduction:

Born-Haber cycle is based on Hess’s law to calculate the lattice enthalpy of ionic compounds and deals with energy changes in formation of ionic compounds.

The energy released when gaseous state ions of unlike charges that are infinitely farther apart combine to form a stable ionic solid is called Lattice energy.  Conversely, the energy required to break the electrostatic force of attraction between the ions of unlike charges in the ionic solid and revert them to gaseous state is also termed as Lattice energy of an ionic solid.

Electron affinity of an atom refers to the energy released when one electron is added to neutral atom in gaseous state.

Hess’s law is applied to calculate the enthalpy changes in a reaction.  According to Hess’s law – “The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes involving in each and every individual steps in the reaction.”  Thus if a reaction involves ‘n’ steps then enthalpy change ΔH° of the reaction is,

  ΔH°= ΔH1°+ΔH2°+ΔH3°....+ΔHn°

Expert Solution & Answer
Check Mark

Answer to Problem 9.118SP

Electron affinity of O is calculated as 844 kJ/mol.

Explanation of Solution

Given data:

    heat of sublimation of Mg = 148 kJ/mol

The first step of Born-Haber cycle involves sublimation of solid Mg into gaseous Mg

    Mg(s) Mg(g) ΔH1° = 148 kJ/mol

The second step of Born-Haber cycle involves dissociation of gaseous O2 into gaseous O atoms.

    12O2(g) O(g) ΔH2° = 12(498.7) kJ/mol

The third step of Born-Haber cycle is ionization of gaseous Mg into gaseous Mg2+ ions.

    Mg(g)Mg(g)+ + e- ΔH3°' = 738.1 kJ/molMg(g)+Mg(g)2+ + e- ΔH3°'' = 1450 kJ/mol

The fourth step of Born-Haber cycle is ionization of gaseous O into gaseous O ions.

    O(g)+ e-O(g)  ΔH4°' = -141 kJ/molO(g)+ e-O(g)2  ΔH4°'' = ? kJ/mol

The fifth and final step of Born-Haber cycle is formation of solid NaCl as a result of binding gaseous Na+ and Cl ions together by electrostatic force of attraction.

    Mg2+(g)+O(g)2 MgO(s) ΔH5° = 3890 kJ/mol

ΔH4°'' corresponding to electron affinity of O ion is calculated by Hess’s law as follows,

    Mg(s) Mg(g) ΔH1° = 148 kJ/mol12O2(g) O(g) ΔH2° = 12(498.7) kJ/molMg(g) Mg(g)+ + e- ΔH3°'  = 738.1 kJ/molMg(g)+ Mg(g)2+ + e- ΔH3°'' = 1450 kJ/molO(g)+ e- O(g)  ΔH4°'  = -141 kJ/molO(g)+ e- O(g)2  ΔH4°''  = ? kJ/molMg2+(g)+O(g)2 MgO(s) ΔH5° = 3890 kJ/mol___________________________________________________Mg(s)+12O2(g) MgO(s) ΔHoverall° = 601.8 kJ/mol

electron affinity of O ion, ΔH4°'' = ΔHoverall°ΔH1°ΔH2°ΔH3°ΔH4°ΔH5° = (601.8 148249.4738.11450+141+3890) = 844 kJ/mol

844 kJ/mol of energy is released (negative sign) when an electron is added to O atom in gaseous state.

Hence electron affinity of O is 844 kJ/mol.

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Chapter 9 Solutions

General Chemistry

Ch. 9.6 - Prob. 1RCCh. 9.7 - Prob. 1PECh. 9.7 - Prob. 2PECh. 9.7 - Prob. 1RCCh. 9.8 - Prob. 1PECh. 9.8 - Prob. 1RCCh. 9.9 - Practice Exercise Draw the Lewis structure for...Ch. 9.9 - Prob. 2PECh. 9.9 - Prob. 3PECh. 9.9 - Prob. 4PECh. 9.9 - Prob. 1RCCh. 9.10 - Prob. 1PECh. 9.10 - Prob. 1RCCh. 9 - Prob. 9.1QPCh. 9 - 9.2 Use the second member of each group from Group...Ch. 9 - Prob. 9.3QPCh. 9 - Prob. 9.4QPCh. 9 - Prob. 9.5QPCh. 9 - Prob. 9.6QPCh. 9 - Prob. 9.7QPCh. 9 - Prob. 9.8QPCh. 9 - Prob. 9.9QPCh. 9 - Prob. 9.10QPCh. 9 - Prob. 9.11QPCh. 9 - Prob. 9.13QPCh. 9 - Prob. 9.14QPCh. 9 - Prob. 9.15QPCh. 9 - Prob. 9.16QPCh. 9 - Prob. 9.17QPCh. 9 - Prob. 9.18QPCh. 9 - Prob. 9.19QPCh. 9 - Prob. 9.20QPCh. 9 - Prob. 9.21QPCh. 9 - 9.22 Explain how the lattice energy of an ionic...Ch. 9 - Prob. 9.23QPCh. 9 - Prob. 9.24QPCh. 9 - Prob. 9.25QPCh. 9 - Prob. 9.26QPCh. 9 - Prob. 9.27QPCh. 9 - Prob. 9.28QPCh. 9 - Prob. 9.29QPCh. 9 - Prob. 9.30QPCh. 9 - Prob. 9.31QPCh. 9 - Prob. 9.33QPCh. 9 - 9.34 Arrange these bonds in order of increasing...Ch. 9 - Prob. 9.35QPCh. 9 - Prob. 9.36QPCh. 9 - Prob. 9.37QPCh. 9 - Prob. 9.38QPCh. 9 - Prob. 9.39QPCh. 9 - Prob. 9.40QPCh. 9 - Prob. 9.41QPCh. 9 - Prob. 9.42QPCh. 9 - Prob. 9.43QPCh. 9 - Prob. 9.44QPCh. 9 - Prob. 9.45QPCh. 9 - Prob. 9.46QPCh. 9 - Prob. 9.47QPCh. 9 - Prob. 9.48QPCh. 9 - Prob. 9.49QPCh. 9 - Prob. 9.50QPCh. 9 - Prob. 9.51QPCh. 9 - Prob. 9.52QPCh. 9 - Prob. 9.53QPCh. 9 - Prob. 9.54QPCh. 9 - Prob. 9.55QPCh. 9 - Prob. 9.56QPCh. 9 - Prob. 9.57QPCh. 9 - Prob. 9.58QPCh. 9 - Prob. 9.59QPCh. 9 - Prob. 9.60QPCh. 9 - Prob. 9.61QPCh. 9 - Prob. 9.62QPCh. 9 - Prob. 9.63QPCh. 9 - Prob. 9.64QPCh. 9 - Prob. 9.65QPCh. 9 - Prob. 9.66QPCh. 9 - Prob. 9.67QPCh. 9 - Prob. 9.68QPCh. 9 - Prob. 9.69QPCh. 9 - Prob. 9.70QPCh. 9 - Prob. 9.71QPCh. 9 - Prob. 9.72QPCh. 9 - Prob. 9.73QPCh. 9 - Prob. 9.74QPCh. 9 - Prob. 9.75QPCh. 9 - Prob. 9.76QPCh. 9 - Prob. 9.77QPCh. 9 - Prob. 9.78QPCh. 9 - Prob. 9.79QPCh. 9 - Prob. 9.80QPCh. 9 - 9.81 Draw reasonable resonance structures for...Ch. 9 - Prob. 9.82QPCh. 9 - Prob. 9.83QPCh. 9 - Prob. 9.84QPCh. 9 - Prob. 9.85QPCh. 9 - Prob. 9.86QPCh. 9 - Prob. 9.87QPCh. 9 - Prob. 9.88QPCh. 9 - Prob. 9.89QPCh. 9 - Prob. 9.90QPCh. 9 - Prob. 9.91QPCh. 9 - Prob. 9.92QPCh. 9 - Prob. 9.93QPCh. 9 - Prob. 9.94QPCh. 9 - Prob. 9.95QPCh. 9 - Prob. 9.96QPCh. 9 - Prob. 9.97QPCh. 9 - Prob. 9.98QPCh. 9 - Prob. 9.99QPCh. 9 - Prob. 9.100QPCh. 9 - Prob. 9.101QPCh. 9 - Prob. 9.102QPCh. 9 - Prob. 9.103QPCh. 9 - Prob. 9.104QPCh. 9 - Prob. 9.105QPCh. 9 - Prob. 9.106QPCh. 9 - Prob. 9.107QPCh. 9 - Prob. 9.108QPCh. 9 - 9.109 Among the common inhaled anesthetics...Ch. 9 - 9.110 Industrially, ammonia is synthesized by the...Ch. 9 - Prob. 9.111QPCh. 9 - Prob. 9.112QPCh. 9 - Prob. 9.113SPCh. 9 - Prob. 9.114SPCh. 9 - Prob. 9.115SPCh. 9 - Prob. 9.116SPCh. 9 - Prob. 9.117SPCh. 9 - Prob. 9.118SP
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