Heat of Hydration Formula with Example

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

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