## Heat of Hydration Formula :

The heat of hydration formula (_{(}_{Kj/mol) }in kilojoule per mole is equal to the product of heat of solution ΔH_{Solution(}_{(}_{Kj/mol) }in kilojoule per mole minus of Lattice energy of the solution ΔH_{latticeenergy }_{(}_{Kj/mol) }in kilojoule per mole .Hence the heat of hydration formula can be written as,

**Heat of hydration _{(Kj/mol )=(}ΔH_{Solution} _{(Kj/mol) (– }ΔH_{latticeenergy (}_{Kj/mol) )}**

Where,

ΔH_{Solution }→ Heat of the solution in KJ/mol

ΔH_{latticeenergy} → Lattice energy of the solution in KJ/mol

### Example 1:

Calculate the heat of hydration of Na^{+} and Cl^{-1} where the heat of hydration of cl- is -400kJ/mol? The sodium chloride lattice enthalpy is ΔHfor→→Na^{+}+Cl^{-1} is 800kJ/mol. To make 2M Nacl the solution heat is +6.0kJ/mol.

### Solution :

Given data

Lattice energy =800kJ/mol

Heat of solution=6.0kJ/mol

Heat of hydration of Cl^{-1}= -400Kj/mol

The formulas as given by

Heat of hydration=(ΔH_{Solution – }ΔH_{latticeenergy } )

=6-800=794

Heat of hydration =-794

Heat of hydration of Na^{+}+Cl^{-1}=-794

Heat of hydration of Na^{+}=-794-(-400)

Heat of hydration of Na^{+}=- 394

### Example 2 :

Calculate the enthalpy of hydration of chloride ions? The lattice enthalpy of solid NaCl is 660 kJmol^{−1 }and the enthalpy of the solution is 4 kJmol^{−1}. If the hydration enthalpy of Na^{+} and Cl^{− }ions are in the ratio of 2:3.5.

### Solution :

Given that

Lattice energy =800kJ/mol

Heat of solution=6.0kJ/mol

Heat of hydration=(ΔH_{Solution – }ΔH_{latticeenergy } )

=-4-660

=-664KJ mol^{-1}

The hydration enthalpy of sodium and chloride ions in the ratio of 2:3.5

So

2x+3.5x=-664

Or

X=-121

The enthalpy of hydration of chloride ion is

-121 ×3.5=-423.5KJ.mol^{-1}