Given Reaction $\quad$ Energy Change (in $\mathbf{k J}$) $\mathrm{Li}(\mathrm{s}) ightarrow…

Given Reaction $\quad$ Energy Change (in $\mathbf{k J}$) $\mathrm{Li}(\mathrm{s}) ightarrow \mathrm{Li}(\mathrm{g})$ : 161 $\mathrm{Li}(\mathrm{g}) ightarrow \mathrm{Li}^{+}(\mathrm{g})$ : 520 $\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) ightarrow \mathrm{F}(\mathrm{g}) \quad 77$ $\mathrm{F}(\mathrm{g})+\mathrm{e}^{-} ightarrow \mathrm{F}^{-}(\mathrm{g})$ : x (Electron gain enthalpy) $\mathrm{Li}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) ightarrow \mathrm{Li} \mathrm{F}(\mathrm{s})-1047$ $\mathrm{Li}(\mathrm{s})+\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) ightarrow \mathrm{Li} \mathrm{F}(\mathrm{s})-617$ Based on data provided, the value of electron gain enthalpy of fluorine would be :
  1. $-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $-350 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $-228 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

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

Asked in: JEE-TOPICTESTS-CHEMISTRY

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