Question
The lattice energy of solid $$NaCl$$ is $$180\,kcal\,mo{l^{ - 1}}$$ and enthalpy of solution is $$1\,kcal\,mo{l^{ - 1}}.$$ If the hydration energies of $$N{a^ + }$$ and $$C{l^ - }\,ions$$ are in the ratio 3 : 2, what is the enthalpy of hydration of sodium $$ion ?$$
A.
$$ - 107.4\,kcal\,mo{l^{ - 1}}$$
B.
$$107.4\,kcal\,mo{l^{ - 1}}$$
C.
$$71.6\,kcal\,mo{l^{ - 1}}$$
D.
$$ - 71.6\,kcal\,mo{l^{ - 1}}$$
Answer :
$$ - 107.4\,kcal\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta {H_{hyd.}} = \Delta {H_{sol.}} - \Delta {H_{lattice}} \cr
& = 1 - 180 = - 179\,kcal\,mo{l^{ - 1}} \cr
& {\text{Then}}\,\,\Delta {H_{hyd.}}\left( {N{a^ + }} \right) + \Delta {H_{hyd.}}\left( {C{l^ - }} \right) = - 179 \cr
& {\text{or}}\,\,\Delta {H_{hyd.}}\left( {N{a^ + }} \right) + \frac{2}{3}\Delta {H_{hyd}} = - 179 \cr
& {\text{or}}\,\,\Delta {H_{hyd.}}\left( {N{a^ + }} \right) = - 107.4\,kcal\,mo{l^{ - 1}} \cr} $$