Question

The lattice enthalpy for lithium fluorides, given the following information: • Enthalpy of sublimation for solid lithium = 161 kJ/mol • First ionization energy for lithium = 520 kJ/mol • F-F bond dissociation energy = 154 kJ/mol • Enthalpy of formation for F(g) = 77 kJ/mol • Electron affinity for fluorine = -328 kJ/mol • Enthalpy of formation for solid lithium fluoride = -617 kJ/mol

Answer 1

Answer:

ANSWER

Given:

ΔsubHLi=155.2kJmol−1

21ΔHF−F=75.3kJmol−1

ΔfHLi=520kJmol−1

ΔegHF=−333kJmol−1

ΔfHLiF=−594.1kJmol−1

For LiF: ΔfHLiF=ΔsubHLi+Δ21HF−F+ΔiHLi+ΔegHF+ΔlHLiF

or −594.1=155.2+75.3+520−

Answer 2

Answer:

The lattice enthalpy for lithium fluoride (LiF) is equal to -1047 KJ/mol.

Explanation:

The formation of lithium fluoride as:

Li⁺ (g)  +  F⁻(g) $\longrightarrow$   LiF (s)

Given, enthalpy of sublimation for solid lithium $\triangle H^o_1= 161 \;KJ/mol$

The 1st ionization energy for lithium , $\triangle H^o_2=520 \;KJ/mol$

The bond dissociation energy of F-F bond $=154 \;KJ/mol$

The bond dissociation energy of F-F bond for half mole, $\triangle H^o_3 = 77\;KJmol^{-1}$

The electron gain enthalpy for F(g) = $\triangle H^o_4=77 \;KJ/mol$

The electron affinity for fluorine ion, $\triangle H^o_4=-328 \;KJ/mol$

The overall enthalpy of formation for solid lithium fluoride,$\triangle H^o= -617 \;KJ/mol$

Consider that ΔH₅⁰ is the enthalpy change for lattice formation.

$\triangle H^o= \triangle H^o_1 +\triangle H^o_2+\triangle H^o_3+ \triangle H^o_4+\triangle H^o_5$

$-\triangle H^o_5= \triangle H^o_1 +\triangle H^o_2+\triangle H^o_3+ \triangle H^o_4-\triangle H^o$

$-\triangle H^o_5 = 161+ 520 +77-328-(-617)$

$-\triangle H^o_5 =1047\;KJmol^{-1}$

ΔH₅⁰  = -1047 KJ/mol

Therefore, the lattice enthalpy for lithium fluoride is equal to -1047 KJ/mol.

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