Question
$${\text{Given}}$$ $${\text{Energy Change}}$$
$${\text{Reaction}}$$ $${\text{(in kJ)}}$$
$$Li\left( s \right) \to Li\left( g \right)$$ $$161$$
$$Li\left( g \right) \to L{i^ + }\left( g \right)$$ $$520$$
$$\frac{1}{2}{F_2}\left( g \right) \to F\left( g \right)$$ $$77$$
$$F\left( g \right) + {e^ - } \to {F^ - }\left( g \right)$$ $${\text{(Electron gain enthalpy)}}$$
$$L{i^ + }\left( g \right) + {F^ - }\left( g \right) \to Li\,\,F\left( s \right)$$ $$ - 1047$$
$$Li\left( s \right) + \frac{1}{2}{F_2}\left( g \right) \to Li\,F\left( s \right)$$ $$ - 617$$
Based on data provided, the value of electron gain enthalpy of fluorine would be :
A.
$$ - 300\,kJ\,mo{l^{ - 1}}$$
B.
$$ - 350\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 328\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 228\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 328\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Applying Hess's}}\,\,{\text{Law}} \cr
& {\Delta _f}{H^ \circ } = {\Delta _{sub}}H + \frac{1}{2}{\Delta _{diss}}H + I.E. + E.A + {\Delta _{lattice}}H \cr
& - 617 = 161 + 520 + 77 + E.A. + \left( { - 1047} \right) \cr
& E.A. = - 617 + 289 = - 328\,kJ\,mo{l^{ - 1}} \cr
& \therefore \,\,{\text{electron affinity of fluorine}} \cr
& = - 328\,kJ\,mo{l^{ - 1}} \cr} $$