Consider Nal → Na+ + - and the following information.
Hlat = -704 kJ/mol
AHhydr of Na+= -410.0 kJ/mol
AHhydr of -= -247 kJ/mol
What is the AHSol of this compound? Use AHsol = -AHlat + AHhydr.
0-867 kJ/mol
|-867.0 kJ/mol
0 47 kJ/mol
0 47.0 kJ/mol

47 kJ·mol⁻¹  

Explanation:

NaI(s) ⟶ Na⁺(aq) + I⁻(aq); ΔH = ?

We can carry out this process in a series of steps.

1. Convert the solid to its gaseous ions.

NaI(s) ⟶ Na⁺(g) + I⁻(g); -ΔH(lat) = 704 kJ

2. Hydrate the Na⁺ ions

Na⁺(g) ⟶ Na⁺(aq); ΔH(hyd) = - 410.0 kJ

3. Hydrate the I⁻ ions

I⁻(g) ⟶ I⁻(aq); ΔH(hyd) = -247

4. Add the equations

Cancel ions that occur on opposite sides of the reaction arrow

                                      ΔH/kJ·mol⁻¹

NaI(s) ⟶ Na⁺(g) + I⁻(g);      +704

Na⁺(g) ⟶ Na⁺(aq);              - 410.0

I⁻(g) ⟶ I⁻(aq);                      -247

NaI(s) ⟶ Na⁺(g) + I⁻(g)           47

The heat of solution of NaI is 47 kJ·mol⁻¹.

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