Chemical Thermodynamics Question 420
Question: The lattice energy of solid $ NaCl $ is $ 180kcalmo{{l}^{-1}} $ and enthalpy of solution is $ 1kcalmo{{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-
Options:
A) $ -107.4kcalmo{{l}^{-1}} $
B) $ 107.4kcalmo{{l}^{-1}} $
C) $ 71.6kcalmo{{l}^{-1}} $
D) $ -71.6kcalmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \Delta {H_{hyd.}}=\Delta {H_{sol.}}-\Delta H_{lattice} $
$ =1-180=-179kcal,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}} $
