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The heat of hydration formula is given by:

Heat of hydration = (ΔH _solution – ΔH _{lattice energy})

Where

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.

Solution:

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 = (ΔH_solution – ΔH_{lattice 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