Answer:
The lattice energy for LiF(s) is -1047 kJ/mol
Explanation:
Given;
Sublimation energy, ΔHs = 166 kJ/mol
bond dissociation energy, BE = +77 kJ/mol
first ionization energy, IE₁ = +520 kJ/mol
electron affinity, EA = -328 kJ/mol
enthalpy of formation, ΔHf = -612 kJ/mol
[tex]L_iF_{(s)}[/tex] ⇄ [tex]Li_{(g)}^+\ \ + \ \ F^-_{(g)}[/tex]
Apply the following equation to calculate the lattice energy for, LiF(s),
Let the lattice energy = U
ΔHf = ΔHs + BE + IE₁ + EA + U
-612 = 166 + 77 + 520 -328 + U
-612 = 435 + U
U = -612 - 435
U = -1047 kJ/mol
Therefore, the lattice energy for LiF(s) is -1047 kJ/mol
The lattice energy of LiF is -1047 KJ/mol.
The lattice energy is the energy is the energy evolved when one mole of the ionic solid is formed from its constituents. The ionic lattice energy depends on the size of the ions.
We have the following information from the question;
Enthalpy of formation of LiF (ΔHf ) = -612 kJ/mol
Heat of sublimation of Li (ΔHs) = +166 kJ/mol
Bond energy of F2 (BE) = +77 kJ/mol
Ionization energy of Li (IE₁) = +520. kJ/mol
Electron affinity of F(g) (EA) = -328 kJ/mol
Lattice energy (U) = ?
The lattice energy can be obtained using the Hess law of constant heat summation as follows;
ΔHf = ΔHs + BE + IE₁ + EA + U
U = ΔHf - [ΔHs + BE + IE₁ + EA ]
U = -612 kJ/mol - [(+166 kJ/mol ) + ( +77 kJ/mol) + (+520. kJ/mol) + ( -328 kJ/mol)]
U =-612 kJ/mol - 435 KJ/mol
U = -1047 KJ/mol
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