A) \[-107.4\text{ }kcal\text{ }mo{{l}^{-1}}\]
B) \[107.4\text{ }kcal\text{ }mo{{l}^{-1}}\]
C) \[71.6\text{ }kcal\text{ }mo{{l}^{-1}}\]
D) \[-71.6\text{ }kcal\text{ }mo{{l}^{-1}}\]
Correct Answer: A
Solution :
[a] \[\Delta {{H}_{hyd.}}=\Delta {{H}_{sol.}}-\Delta {{H}_{lattice}}\] \[=1-180=-179\text{ }kcal\,mo{{l}^{-1}}\] Then \[\Delta {{H}_{hyd.}}=(N{{a}^{+}})+\Delta {{H}_{hyd.}}(C{{l}^{-}})=-179\] or \[\Delta {{H}_{hyd.}}=(N{{a}^{+}})+\frac{2}{3}\Delta {{H}_{hyd.}}=-179\] or \[\Delta {{H}_{hyd.}}=(N{{a}^{+}})=-107.4\,kcal\,mo{{l}^{-1}}\]You need to login to perform this action.
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