Part 2 ΔH Enthalpy Changes continued: Enthalpies of solution, enthalpies of Ion hydration, Hess's Law cycles continued

2.1d–e The Hess's Law cycle of the energy changes when a solid dissolves in a solvent e.g. water – Calculations involving Enthalpy of hydration of ions, enthalpy of solution, lattice enthalpy

Page introduction

The Hess's Law cycle connecting lattice enthalpy, enthalpies of hydration of ions and the enthalpy of solution is constructed and how to use it in calculations.

A discussion of four cases of dissolving/insolubility are discussed with the aid of enthalpy level diagrams using the Hess's Law cycle for dissolving previously described.

2.1 continued – The Hess's Law cycle of the energy changes when a solid dissolves

See Part 2.2 for Born–Haber Cycle Calculations

Section 2.1d Examples of 'dissolving' enthalpy cycles using Hess's Law

Example 1 Dissolving the salt sodium chloride

From the processes (i), (ii) and (iii) you can put them together in a Hess's Law enthalpy cycle to determine various enthalpy unknowns concerning dissolving an ionic compound in water.

(i) = ΔHθsolution, 298K(ionic substance) = ?? kJmol–1
(exothermic or endothermic) (i) ionic compound(s) + aq cations(aq)  +  anions(aq)
(ii) = ΔHθLE(ionic compound(s)) lattice enthalpy of ionic compound	(endothermic)	cations(g) + anions(g)	(exothermic)	(iii) = ΔHθhyd(cations(aq)) + ΔHθhyd(anions(aq)) sum of the enthalpies of hydration of the cations and anions
Applying Hess's Law: (i) = (ii) + (iii), so in general : ΔHθsolution, 298K(ionic compound) = = ΔHθLE(compound(s)) + ΔHθhyd(cations(g)) + ΔHθhyd(anions(g)) kJmol–1 = ΔHθLE(NaCl(s)) + ΔHθhyd(Na+(g)) + ΔHθhyd(Cl-(g)) kJmol–1

From above the processes (i), (ii) and (iii) are now put together in a Hess's Law cycle for sodium chloride. Watch the ΔH signs, e.g. here the ΔH_LE sign is +ve

(i) = ΔHθsolution, 298K(sodium chloride) = +4 kJmol–1
(i) NaCl(s) + aq Na+(aq)  +  Cl–(aq)
(ii) = ΔHθLE(NaCl(s)) lattice enthalpy of sodium chloride			(iii) = ΔHθhyd(Na+(aq)) + ΔHθhyd(Cl–(aq)) sum of the enthalpies of hydration of the sodium ion and chloride ion
Na+(g)  +  Cl–(g)
Applying Hess's Law: (i) = (ii) + (iii) ΔHθsolution, 298K(sodium chloride) = = ΔHθLE(NaCl(s)) + ΔHθhyd(Na+(g)) + ΔHθhyd(Cl–(g)) = 774) + (–406) + (–364) = +774 –406 –364 = +4 kJmol–1

Example 2 Dissolving the alkali potassium hydroxide

(i) = ΔHθsolution, 298K(potassium hydroxide) = ? kJmol–1
(i) KOH(s) + aq K+(aq)  +  OH–(aq)
(ii) = ΔHθLE(KOH(s)) lattice enthalpy of potassium hydroxide			(iii) = ΔHθhyd(K+(aq)) + ΔHθhyd(OH–(aq)) sum of the enthalpies of hydration of the sodium ion and chloride ion
K+(g)  +  OH–(g)
Applying Hess's Law: (i) = (ii) + (iii) ΔHθsolution, 298K(potassium hydroxide) = = ΔHθLE(KOH(s)) + ΔHθhyd(K+(g)) + ΔHθhyd(OH–(g)) kJmol–1

Example 3 Dissolving the salt aluminium fluoride

(i) = ΔHθsolution, 298K(aluminium fluoride) = ? kJmol–1
(i) AlF3(s) + aq Al3+(aq)  +  3F–(aq)
(ii) = ΔHθLE(AlF3(s)) lattice enthalpy of sodium chloride			(iii) = ΔHθhyd(Al3+(aq)) + 3 x ΔHθhyd(F–(aq)) sum of the enthalpies of hydration of the sodium ion and chloride ion
Al3+(g)  +  3F–(g)
Applying Hess's Law: (i) = (ii) + (iii)                           ΔHθsolution, 298K(aluminium fluoride) = ΔHθLE(AlF3(s)) + ΔHθhyd(Al3+(g)) + {3 x ΔHθhyd(F–(g))} kJmol–1

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Section 2.1e Examples of explaining salt solubility or insolubility

In the case of salts/binary compounds etc. that do not dissolve, the bonding forces are too strong e.g. iron(III) oxide, aluminium oxide etc. where you have two highly charged ions attracting each other. more on this below in which entropy changes are mentioned, but entropy as a concept is dealt with in detail on Part 3 Entropy and Free Energy changes.

Four generalised cases are discussed below but using enthalpy level diagrams which are helpful in understanding the situation rather then the enthalpy cycle diagrams, which are better for doing calculations in my opinion.

The enthalpy of hydration is the enthalpy of solvation when water is the solvent.

Water is a very polar solvent and is one of the best solvents for ionic compounds.

In example 4. a non–polar solvent is considered, so the term enthalpy of solvation applies rather than enthalpy of hydration.

	Ex 2.1e1: In example 1. The total enthalpies of hydration of the ions is very high i.e. very exothermic, the lattice enthalpy is moderately high (moderately endothermic) and the enthalpy of solution is moderately exothermic. Under these 'energetics' circumstances you would expect the solute to be soluble because the exothermic enthalpies of hydration far outweigh the endothermic lattice enthalpy change. Note again the ΔHLE sign is +ve
	Ex 2.1e2: In example 2. the total enthalpy of hydration is far short of overriding the endothermic lattice enthalpy and leads to an endothermic enthalpy of solution. This substance might well be ~insoluble but if on dissolving the entropy change is positive enough, it might well still dissolve. In this case of a slightly endothermic enthalpy of solution the entropy change becomes most important (see the three points below). So the salt may or may not dissolve.
Some general points then ... The dissolving process involves an increase in entropy favouring dissolving – the solution–mixture is more disordered (more possible arrangements of the particles) compared to the pure liquid solvent and the highly ordered crystal lattice. However, when the ions become hydrated there is a decrease in entropy, NOT favouring the dissolving, which counts against a substance dissolving. This decrease in entropy is due to the orientation of the water molecules when they associate with the cations and anions i.e. a decrease in the possible ways the water molecules can be arranged. Therefore, in these 'marginal' cases, the compound may be soluble or insoluble depending on the subtle nature of the entropy changes! However, it is worth noting, that if the solute is NOT ionic, then there is no lattice enthalpy to overcome so small covalent molecules like iodine will readily dissolve in solvents like hexane or ethanol, if not very soluble in highly polar water.
	Ex 2.1e3:   In example 3. the total enthalpy of hydration is quite low and far short of overriding the quite high endothermic lattice enthalpy and leads to quite an endothermic enthalpy of solution. This substance is quite likely to be insoluble.
	Ex 2.1e4:   In example 4. we consider a non–polar solvent which will have little association with ions (unlike the highly polar water). Consequently the enthalpy of solvation is very low and far short of overriding the high endothermic lattice enthalpy and leads to a very endothermic enthalpy of solution. This substance is very likely to be insoluble. Even the solubility of salts in organic polar molecules like ethanol is usually quite low.

See Part 2.2 for Born–Haber Cycle Calculations

Energetics-Thermochemistry-Thermodynamics Notes INDEX

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QUICK INDEX for Energetics: INDEX of ALL advanced level pages on thermochemistry and thermodynamics Part 1a–b ΔH Enthalpy Changes 1.1 Advanced Introduction to enthalpy changes of reaction, formation, combustion etc. : 1.2a & 1.2b(i)–(iii) Thermochemistry – Hess's Law and Enthalpy Calculations – reaction, combustion, formation etc. : 1.2b(iv) Enthalpy of reaction from bond enthalpy calculations  : 1.3a–b Experimental methods for determining enthalpy changes and treatment of results and calculations : 1.4 Some enthalpy data patterns : 1.4a The combustion of linear alkanes and linear aliphatic alcohols : 1.4b Some patterns in Bond Enthalpies and Bond Length : 1.4c Enthalpies of Neutralisation : 1.4d Enthalpies of Hydrogenation of unsaturated hydrocarbons and evidence of aromatic ring structure in benzene : Extra Q page A set of practice enthalpy calculations with worked out answers ** Part 2 ΔH Enthalpies of ion hydration, solution, atomisation, lattice energy, electron affinity and the Born–Haber cycle : 2.1a–c What happens when a salt dissolves in water and why? : 2.1d–e Enthalpy cycles involving a salt dissolving : 2.2a–c The Born–Haber Cycle *** Part 3 ΔS Entropy and ΔG Free Energy Changes : 3.1a–g Introduction to Entropy : 3.2 Examples of entropy values and comments * 3.3a ΔS, Entropy and change of state : 3.3b ΔS, Entropy changes and the feasibility of a chemical change : 3.4a–d More on ΔG, free energy changes, feasibility and applications : 3.5 Calculating Equilibrium Constants from ΔG the free energy change : 3.6 Kinetic stability versus thermodynamic feasibility - can a chemical reaction happen? and will it happen? Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's Chemistry revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Advanced level pre-university/college chemistry revision notes and key points on energetics, thermochemistry, thermodynamics including enthalpy, entropy and free energy changes: All suitable for chemistry students taking the WJEC advanced A level chemistry, CCEA advanced A level chemistry, Cambridge CIE advanced A level chemistry, AQA advanced A level chemistry, Edexcel advanced A level chemistry, OCR advanced A level chemistry, Salters advanced A level chemistry, IB advanced level chemistry, US grade 11-12 K12 AP Honors chemistry courses INDEX of all my notes on thermochemistry and thermodynamics