binary ionic compounds · binary ionic compounds chapter 8 part ii. energetics of ionic bonding ......
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Binary Ionic
Compounds
Chapter 8 Part II
Energetics of Ionic Bonding
• There must be a third piece to the puzzle.
• What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.
Energetics of Ionic Bonding
Lattice Energy • This third piece of the puzzle is the lattice energy: The energy required to completely separate a mole of a solid ionic compound into its gaseous ions.
• The energy associated with electrostatic interactions is governed by Coulomb’s law:
F = K Q1Q2
d
Lattice Energy • Lattice energy, then, increases with the charge on the ions.
• It also increases with decreasing size of ions.
Energetics of Ionic Bonding
By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.
Energetics of Ionic Bonding
Energetics of Ionic Bonding
• These phenomena also helps explain the “octet rule.”
Energetics of Ionic Bonding
• Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies.
• Bond-dissociation energy is the energy required to break one mole of a particular type of covalent bond in a gas-phase compound.
• Energies of some bonds can differ from compound to compound, so we use an average bond energy.
Bond Energy
Bond Energy
The H—H bond
energy is precisely
known …
… while the O—H bond
energies for the two bonds
in H2O are different.
Average Bond Enthalpies
• This table lists the average bond enthalpies for many different types of bonds.
• Average bond enthalpies are positive, because bond breaking is an endothermic process.
Average Bond Enthalpies NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the
C—H bond in chloroform, CHCl3.
Average Bond Enthalpies
Enthalpies of Reaction • Yet another way to estimate H
for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed.
• In other words, Hrxn = (bond enthalpies of bonds broken)
(bond enthalpies of bonds formed)
For the reaction N2(g) + 2 H2(g) N2H4(g) to occur …
Calculations Involving Bond Energies
… plus 2(436 kJ),
to break bonds of
the reactants.
… we must supply
946 kJ …
When the bonds of the
product form, 163 kJ plus
4(389 kJ) of energy is
liberated.
∆ H = (+946 kJ) + 2(+436 kJ) + (–163 kJ) + 4(–389 kJ)
Enthalpies of Reaction CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
In this example, one C—H bond and one Cl—Cl bond are broken; one C—Cl and one H—Cl bond are formed.
18
Bond
Average H/mol
Bond
Average H/mol
C-H 413 Cl-Cl 242
H-Cl 431 C-Cl 328
C-C 348 C=C 614
Hrxn ≈ (Hbonds broken) - (Hbonds
formed) Hrxn ≈ [(1(413) + 1(242)] –
[1(328) + 1(431)] Hrxn ≈ -104 kJ/mol
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