Heat of Hydration Formula :
The heat of hydration formula ((Kj/mol) in kilojoule per mole is equal to the product of heat of solution ΔHSolution((Kj/mol) in kilojoule per mole minus of Lattice energy of the solution ΔHlatticeenergy (Kj/mol) in kilojoule per mole .Hence the heat of hydration formula can be written as,
Heat of hydration(Kj/mol )=(ΔHSolution (Kj/mol) (– ΔHlatticeenergy (Kj/mol) )
Where,
ΔHSolution → Heat of the solution in KJ/mol
ΔHlatticeenergy → 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=(ΔHSolution – ΔHlatticeenergy )
=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=(ΔHSolution – ΔHlatticeenergy )
=-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