Question

Calculate the hydration enthalpy of ions for $NaCl\left(s\right)$ from the following data.
${\Delta }_{lattice}H=+788kJmo{l}^{-1}$
${\Delta }_{sol}H=+4kJmo{l}^{-1}$

• A. $-784kJmo{l}^{-1}$
• B. $-1084kJmo{l}^{-1}$
• C. $+784kJmo{l}^{-1}$
• D. $+1084kJmo{l}^{-1}$

Answer 1

The correct option is A $-784kJmo{l}^{-1}$

We know that,
${\Delta }_{soln}H={\Delta }_{hyd}H+{\Delta }_{lattice}H$
but given that,
${\Delta }_{lattice}H=+788kJmo{l}^{-1}$
${\Delta }_{soln}H=+4kJmo{l}^{-1}$

putting the above values in the formula we get,
$+4kJmo{l}^{-1}={\Delta }_{hyd}H+788kJmo{l}^{-1}$
${\Delta }_{hyd}H=-784kJmo{l}^{-1}$


Theory:

Lattice enthalpy ${\Delta }_{Lattice}H$:
Energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions.

Standard lattice enthalpy ${\Delta }_{Lattice}{H}^o$:
Energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions at unit pressure $\left(1bar\right)$ and temperature of $298K$.

Standard enthalpy of hydration ${\Delta }_{hyd}{H}^o$Conditions specified = $1bar$ and $298K$
Enthalpy change when one mole of anhydrous or partially hydrated salt combines with the requisite amount of water to form a new hydrated stable salt at unit pressure $\left(1bar\right)$ and $298K$.

Integral enthalpy of solution/enthalpy of dissolution:
Enthalpy change when one mole of the solute is dissolved in a definite amount of the solvent to make a solution.
${\Delta }_{soln}H={\Delta }_{hyd}H+{\Delta }_{lattice}H$