EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 8220101425812
Author: DECOSTE
Publisher: Cengage Learning US
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Chapter 13, Problem 142CP
Interpretation Introduction

Interpretation: The possible structures and enthalpy for MaXb should be determined.

Concept Introduction:

The enthalpy is defined as the product of pressure and volume added up with the system’s internal energy. The change in enthalpy is defined as the heat absorbed or released. The bond energy is also known as the mean bond enthalpy or the average bond enthalpy. It is the measure of the bond strength depends on the chemical bond.

Expert Solution & Answer
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Answer to Problem 142CP

  M2X has the more negative enthalpy formation. The charges of the ions in M2X are M+ and X2-

Explanation of Solution

Given information:

Successive ionization energies of M: 480, 4750

Successive electron affinity values for X: -175, 920.

Enthalpy of sublimation for M(s) —> M (g): 110.

Bond energy of X2:250

Lattice energies are MX=-1200kJ/mol;MX2=-3500kJ/mol;M2X=-3600kJ/mol

To determine all the possible structures. There are four possible structures for the compound MaXb . The compound is MX that can be formed by two different ways M2X and MX2 . To determine ΔH of formation of all the compounds is given below,

The equation is formed as,

  M(s)M(g)ΔH=110kJM(g)M+(g)+e-ΔH=480kJM+(g)M2+(g)+e-ΔH=4750kJ12X2(g)X(g)ΔH=12(250kJ)X(g)+e-X-(g)ΔH=-175kJX-(g)+e-X2-(g)ΔH=920kJM2+(g)+X2-(g)MX(g)ΔH=-4800kJ

For MX composed of M2+andX2-,M2+(s)+X2+(g)MX(s):

  ΔHf=110+480+4750+12(250)1751200920ΔHf=1410kJ

For MX composed of M+andX-,2M(s)+12X2(g)MX(s):

  ΔHf=110+480+12(250)1751200ΔHf=660kJ

For MX composed of M+andX2-,2M(s)+12X2(g)M2X(s):

  ΔHf=2(110)+2(480)+12(250)1759203600ΔHf=1550kJ

For MX composed of M2+andX-,M(s)+X2(g)M2X(s):

  ΔHf=110+480++4750+250+2(17)53500ΔHf=1740kJ

The compound with the most negative enthalpy values are most likely to form. Only M+X-and(M+)2X2- are possible because they have negative enthalpies of formation. Since M2X has the more negative enthalpy op formation, it is most likely to form. Thus, the charges of the ions in M2X are M+ and X2-

Conclusion

  M2X has the more negative enthalpy formation. The charges of the ions in M2X are M+ and X2-

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Consider an ionic compound, MX, , composed of generic metal M and generic gaseous halogen X. • The enthalpy of formation of MX, is AH; = -813 kJ/mol. t. • The enthalpy of sublimation of M is AHsub = 181 kJ/mol. • The first, second, and third ionization energies of M are IE¡ = 589 kJ/mol, IE2 = 1609 kJ/mol, and IE3 = 2571 kJ/mol. • The electron affinity of X is AHEA = -321 kJ/mol. (Refer to the hint). • The bond energy of X, is BE = 179 kJ/mol. Determine the lattice energy of MX,. AHjattice = kJ/mol
Consider an ionic compound, MX,3, composed of generic metal M and generic gaseous halogen X. • The enthalpy of formation of MX, is AH; = -763 kJ/mol. • The enthalpy of sublimation of M is AHgub 169 kJ/mol. • The first, second, and third ionization energies of M are IE1 619 kJ/mol, IE2 = 1667 kJ/mol, and IE3 = 2525 kJ/mol. • The electron affinity of X is AHEA -307 kJ/mol. (Refer to the hint). • The bond energy of X, is BE = 181 kJ/mol. Determine the lattice energy of MX,. AHjattice kJ/mol II
Consider an ionic compound, MX,, composed of generic metal M and generic, gaseous halogen X. • The enthalpy of formation of MX, is AH; -909 kJ/mol. • The enthalpy of sublimation of M is AHsub = 143 kJ/mol. • The first and second ionization energies of M are IE1 :759 kJ/mol and IE2 = 1407 kJ/mol. • The electron affinity of X is AHEA = -339 kJ/mol. (Refer to the hint). • The bond energy of X, is BE = 177 kJ/mol. Determine the lattice energy of MX,. ΔΗ, kJ/mol lattice * TOOLS х10

Chapter 13 Solutions

EBK CHEMICAL PRINCIPLES

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