Enthalpy of Hydration (Oxford AQA International A Level Chemistry)
Revision Note
Written by: Philippa Platt
Reviewed by: Stewart Hird
Enthalpy of Hydration Calculations
Enthalpy of solution
The standard enthalpy change of solution (ΔHsolꝋ) is the enthalpy change when one mole of an ionic substance dissolves in sufficient water to form an infinitely dilute solution
The symbol (aq) is used to show that the solid is dissolved in sufficient water
For example, the enthalpy changes of solution for potassium chloride are described by the following equations:
KCl (s) + aq → KCl (aq)
OR
KCl (s) + aq → K+(aq) + Cl-(aq)
ΔHsolꝋ can be exothermic (negative) or endothermic (positive)
Enthalpy of hydration
The standard enthalpy change of hydration (ΔHhydꝋ) is the enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form an infinitely dilute solution
For example, the enthalpy change of hydration for magnesium ions is described by the following equation:
Mg2+(g) + aq → Mg2+(aq)
Hydration enthalpies are the measure of the energy that is released when there is an attraction formed between the ions and water molecules
Hydration enthalpies are exothermic
When an ionic solid dissolves in water, positive and negative ions are formed
Water is a polar molecule with a δ- oxygen (O) atom and δ+ hydrogen (H) atoms which will form ion-dipole attractions with the ions present in the solution
The oxygen atom in water will be attracted to the positive ions and the hydrogen atoms will be attracted to the negative ions
Hydration of cations and anions
How are enthalpy of solution and hydration enthalpies related?
Enthalpy of solution = - lattice formation enthalpy + hydration enthalpy
OR
Enthalpy of solution = lattice dissociation enthalpy + hydration enthalpy
The hydration enthalpy is the sum of the hydration enthalpies of each ion
If there is more than one cation or anion, such as in MgCl2, then you must multiply by the appropriate coefficient for that ion
Calculations
Questions in this topic typically ask you to calculate the hydration enthalpy of one of the ions, given the lattice enthalpy, enthalpy of solution and hydration enthalpy of the other ion
This can be done by constructing an appropriate Hess's Law cycle to find the unknown energy value
The energy cycle above shows that there are two routes to go from the gaseous ions to the ions in an aqueous solution
According to Hess’s law, the enthalpy change for both routes is the same, such that:
ΔHsolꝋ = - ΔHlattꝋ + ΔHhydꝋ
ΔHhydꝋ = ΔHlattꝋ + ΔHsolꝋ
Each ion will have its own enthalpy change of hydration, ΔHhydꝋ, which will need to be taken into account during calculations
The total ΔHhydꝋ is found by adding the ΔHhydꝋ values of both anions and cations together
Worked Example
Using the following data, calculate the enthalpy of solution, ΔHsolꝋ for NaCl.
ΔHlattꝋ [NaCl] = -774 kJ mol-1
ΔHhydꝋ [Na+] = -406 kJ mol-1
ΔHhydꝋ [Cl-] = -364 kJ mol-1
Answer:
Step 1: Draw the energy cycle
Step 2: Substitute the values to find ΔHsol [NaCl]
ΔHsolꝋ = - ΔHlattꝋ + ΔHhydꝋ
ΔHsolꝋ[NaCl] = -(-774) + (-406) + (-364) = +4 kJ mol-1
Alternative diagram
Worked Example
Construct an energy cycle to calculate the ΔHhydꝋof magnesium ions in magnesium chloride, given the following data:
ΔHlattꝋ[MgCl2] = -2592 kJ mol-1
ΔHsolꝋ[MgCl2] = -55 kJ mol-1
ΔHhydꝋ[Cl-] = -364 kJ mol-1
Answer:
Step 1: Draw an energy cycle:
Step 2: Substitute the values to find ΔHhydꝋ [Mg2+]
ΔHsolꝋ = - ΔHlattꝋ + ΔHhydꝋ
ΔHhydꝋ = ΔHlattꝋ + ΔHsolꝋ
ΔHhydꝋ[Mg2+] = (-2592) + (-55) - (2 x -364) = -1921 kJ mol-1
Alternative diagram
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