The lattice dissociation energy of solid is . The dissolution of in water is experimentally endother — Thermodynamics and Thermochemistry Chemistry Question
Question
The lattice dissociation energy of solid $NaCl$ is $180 \text{ kcal mol}^{-1}$. The dissolution of $NaCl(s)$ in water is experimentally endothermic to the extent of $1 \text{ kcal mol}^{-1}$. If the individual hydration energies of $Na^+$ and $Cl^-$ are precisely in the ratio of $6:5$, what is the absolute magnitude of the enthalpy of hydration of the $Na^+$ ion alone in $\text{kcal mol}^{-1}$?
💡 Solution & Explanation
The enthalpy of solution is given by $\Delta H_{sol} = \Delta H_{lattice\_dissoc} + \Delta H_{hyd}(Na^+) + \Delta H_{hyd}(Cl^-)$. Given $\Delta H_{sol} = +1 \text{ kcal/mol}$ and Lattice $= 180 \text{ kcal/mol}$. Thus, $1 = 180 + \Delta H_{hyd}(total) \Rightarrow \Delta H_{hyd}(total) = -179 \text{ kcal/mol}$. Let the separated hydration energies be $6x$ and $5x$. $6x + 5x = -179 \Rightarrow 11x = -179 \Rightarrow x = -16.273$. The specific hydration energy of $Na^+$ is $6x = 6 \times (-16.273) = -97.63 \text{ kcal/mol}$, rounding to magnitude $97.6$.