In simple terms
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Dissolving: An Energetic Tug-of-War
Dissolving an ionic salt is an energy battle between breaking apart the crystal lattice and forming new bonds with water. The overall energy change, the enthalpy of solution, tells us whether the process releases or absorbs heat.
Imagine a tightly-knit group of friends holding hands at a party (the ionic lattice). To get them to mingle with the wider crowd (water molecules), you first need to spend energy pulling their hands apart. Once separated, energy is released as each individual is happily surrounded by new acquaintances. Whether the overall party atmosphere gets 'hotter' (exothermic) or 'colder' (endothermic) depends on whether more energy was released by mingling than was spent breaking them up.
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The overall energy change of dissolving (ΔH_sol) is found using a Hess cycle, linking the energy to break the lattice (+ΔH_latt) and the energy released when ions hydrate (ΣΔH_hyd).
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Dissolving is favoured if the energy released by hydration is greater than the energy needed to break the lattice, making the overall process exothermic.
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Hydration enthalpy (ΔH_hyd) is more exothermic for ions with a higher charge and smaller radius, as this creates a stronger attraction to water molecules.
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By combining Born-Haber cycles (for lattice energy) and hydration cycles, we can construct a full thermodynamic picture to predict and explain solubility.
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1. Defining the Key Enthalpy Changes
To understand the energetics of dissolving, we must first define two crucial enthalpy changes: the enthalpy change of solution and the enthalpy change of hydration. Both are defined under standard conditions of 298 K and 100 kPa.
Standard Enthalpy Change of Solution (): This is the enthalpy change when one mole of a substance dissolves in a large excess of solvent to form a solution that is 'infinitely dilute'. For sodium chloride, the equation is: . This value can be exothermic (negative) or endothermic (positive).
Standard Enthalpy Change of Hydration (): This is the enthalpy change when one mole of isolated gaseous ions is dissolved in water to form hydrated aqueous ions. For a sodium ion, the equation is: . This process is always exothermic because new, stable ion-dipole forces are formed between the ion and polar water molecules.
2. The Dissolving Process: An Enthalpy Cycle
We can't directly measure the energy change of breaking a lattice and then hydrating the ions separately. However, using Hess's Law, we can construct an energy cycle that connects these processes. The overall process of dissolving an ionic solid can be thought of as a two-step hypothetical route:
- Overcoming the lattice enthalpy to break the solid into gaseous ions.
- Hydrating these gaseous ions to form aqueous ions.
In this cycle, the direct route is the enthalpy of solution. The indirect route involves breaking the lattice (an endothermic process, hence the negative sign in front of the exothermic ) and then hydrating the resulting gaseous ions (an exothermic process). The sum of the enthalpy changes in the indirect route must equal the enthalpy change of the direct route.
Examiners love to test the sign of the lattice enthalpy in this cycle. Standard lattice enthalpy () is defined for the formation of 1 mole of the lattice from gaseous ions, which is exothermic. The cycle for solution involves breaking the lattice, which is the reverse process and therefore endothermic. Always use the value in your calculation.
3. Factors Affecting Enthalpy of Hydration
The magnitude of the enthalpy of hydration is determined by the strength of the electrostatic attraction between the ion and the polar water molecules. This strength is governed by the ion's charge density.
Ionic Charge: The greater the charge on an ion, the stronger its attraction to the partial charges on water molecules ( on oxygen, on hydrogen). Therefore, becomes more exothermic (more negative). For example, for Mg²⁺(g) is significantly more exothermic than for Na⁺(g).
Ionic Radius: For ions with the same charge, a smaller ionic radius means the charge is concentrated in a smaller volume. This higher charge density leads to a stronger electric field and a more powerful attraction to water molecules, making more exothermic. For example, for Li⁺(g) is more exothermic than for K⁺(g).
Worked examples
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Calculate the standard enthalpy change of solution for magnesium chloride, MgCl₂, using the data below.
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Step 1: Identify the components of the Hess's Law cycle. The equation is .
The lattice enthalpy of silver chloride, AgCl, is . The sum of the hydration enthalpies of its ions, , is .
(a) Calculate the enthalpy of solution for AgCl. (b) Use your answer and the data for NaCl () to comment on the relative solubility of the two salts.
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(a) Calculation of
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What is the standard enthalpy change of solution, ?
The enthalpy change when one mole of an ionic solid dissolves in a sufficient amount of solvent to form an infinitely dilute solution, under standard conditions (298 K and 100 kPa).
Key takeaways
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Standard Enthalpy Change of Solution (): This is the enthalpy change when one mole of a substance dissolves in a large excess of solvent to form a solution that is 'infinitely dilute'. For sodium chloride, the equation is: . This value can be exothermic (negative) or endothermic (positive).
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Standard Enthalpy Change of Hydration (): This is the enthalpy change when one mole of isolated gaseous ions is dissolved in water to form hydrated aqueous ions. For a sodium ion, the equation is: . This process is always exothermic because new, stable ion-dipole forces are formed between the ion and polar water molecules.
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Practice Questions: Enthalpies of Solution and Hydration
Practice Questions: Enthalpies of Solution and Hydration
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