Question

$Li(s) \to Li(g)$	161
$Li(g) \to Li^{+}(g)$	520
$\frac{1}{2} F_2 (g) \to F(g)$	77
$F(g) + e^{-} \to F-(g)$	(Electron gain enthalpy)
$Li^{+} (g) + F^{-}(g) \to Li F(s)$	-1047
$Li(s) + \frac{1}{2} F_2(g) \to LiF(s)$	-617

Based on data provided, the value of electron gain enthalpy of fluorine would be :

• A. $- 300 \, kJ \, mol^{-1}$
• B. $- 350 \, kJ \, mol^{-1}$
• C. $- 328 \, kJ \, mol^{-1}$
• D. $- 228 \, kJ \, mol^{-1}$

Answer 1

Answer: (C)

Applying Hess's Law
$\Delta_{f} H^{\circ}=\Delta_{\text {sub }} H+\frac{1}{2} \Delta_{\text {diss }} H+I . E+E . A+\Delta_{\text {lattice }} H$
$-617=161+520+77+$ E.A. $+(-1047)$
E. $A .=-617+289=-328 \,kJ \,mol ^{-1}$
$\therefore $ electron affinity of fluorine $=-328\, kJ\, mol ^{-1}$