The lattice dissociation energy of solid is 180 kcal/mol. The dissolution of in water is experimenta — Thermodynamics and Thermochemistry Chemistry Question
Question
The lattice dissociation energy of solid $NaCl$ is 180 kcal/mol. The dissolution of $NaCl(s)$ in water is experimentally endothermic (+1 kcal/mol). If the hydration energies of $Na^+$ and $Cl^-$ are exactly in a 6:5 ratio, what is the hydration enthalpy of $Na^+$ alone?
Answer: A
💡 Solution & Explanation
Enthalpy of solution $\Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hyd}(total)$. $+1 = 180 + \Delta H_{hyd}(total) \Rightarrow \Delta H_{hyd}(total) = -179$ kcal/mol. Given ratio 6:5, let energies be $6x$ and $5x$. $11x = -179 \Rightarrow x = -16.27$. The hydration energy of $Na^+$ is $6x = 6(-16.27) = -97.6$ kcal/mol.
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