Question

The lattice energy of solid $NaCl$ is 180 kcal/mol. The dissolution of the solid in water in the form of ions in endothermic to extent of a 1 kcal per mole. If the hydration energy of $N{a}^{+}$ ion and $C{l}^{-}$ ion are in the ratio 6:5 then what is the enthalpy of hydration of $N{a}^{+}$ ion.

• A. $-97.5$ kcal/mol
• B. $-9.75$ kcal/mol
  
• C. $-975$ kcal/mol
  
• D. $-9750$ kcal/mol
  

Answer 1

Answer: (A)

$-97.5$ kcal/mol

$\text{Enthalpy of hydration + Energy supplied = Enthalpy of lattice}$

$\text{So enthalpy of hydration = 180 - 1 = 179 kcal/mol}$

Let $6y$ is the energy of hydration of the $N{a}^{+}$

So, $6y+5y=179$

$y=16.27$

So, $6y=97.5$ kcal/mol

The sign will be -ve as this energy is released during the process.