Answer : The molar enthalpy of formation of LiF will be -607 KJ.
Explanation : In the given reaction of
[tex]Li _{(s)} + \frac{1}{2} F_{2} _{(g)} ----\ \textgreater \ LiF_{(s)}[/tex]
we have to use hess's law of summation in this reaction, which is attached in the attachment;
we have got,
[tex]Li ^{+} + F^{-} ----\ \textgreater \ LiF [/tex] ΔU = -1037 KJ/mol
[tex] Li_{s} ----\ \textgreater \ Li_{g} [/tex] [tex]ΔH_{s}[/tex] = 159.37 KJ/mol
[tex] Li_{g} ----\ \textgreater \ Li^{+} [/tex] IE = 520 KJ/mol
[tex] \frac{1}{2} F_{2} _{(g)}----\ \textgreater \ F_{g} [/tex] [tex] D_{0} [/tex] = 78.99 KJ/mol
[tex] F_{(g)} ----\ \textgreater \ F^{-} _{(g)} [/tex] EA = -328.0 KJ/mol
Hence ΔH(formation) = ΔU+[tex]ΔH_{s}[/tex]+[tex] D_{0} [/tex]+EA+IE
ThereforeΔH(formation) = -1037 + 159.37 + 78.99 + -328 + 520
= -607 KJ/mol
So, the enthalpy of formation of LiF will be as -607 KJ/mol
