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The heat of hydration formula is given by:

Heat of hydration = (ΔH solution – ΔH lattice energy)


ΔH solution  = Heat of the solution

ΔH lattice energy = Lattice energy of the solution

Example 1

The sodium chloride lattice enthalpy is ΔH for NaCl →→ Na+ + Cl is 700 kJ/mol. To make 1M NaCl the solution heat is +5.0kJ/mol. Determine the heat of hydration of Na+ and Cl-, where the heat of hydration of Cl- is -300kJ/mol.


Given data

Lattice energy = 700 kJ/mol

Heat of solution = 5.0kJ/mol

Heat of hydration of Cl = -300kJ/mol

Substitute the values in the given formula

Heat of hydration = (ΔHsolution – ΔHlattice energy)

                                       = 5 – 700

Therefore, Heat of hydration = -695

Heat of hydration of Na+ + Cl = -695

Heat of hydration of Na+ = -695 – (-300)

Therefore, Heat of hydration of Na+ = -395

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