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Part 2 ΔH Enthalpy Changes contd. – enthalpies of solution, enthalpies of Ion hydration, Hess's Law cycles

2.1 continued – 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

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.

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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 above the processes (i), (ii) and (iii) put together in a Hess's Law enthalpy cycle for dissolving an ionic compound.

(i) ΔHθsolution,298K(ionic substance) = ?? kJmol–1
(exothermic or endothermic)

(i) ionic compound(s) + aq (c) doc b cations(aq)  +  anions(aq)

(ii) =

–ΔHθLE(ionic compound(s))

– lattice enthalpy of ionic compound

 

(c) doc b

(endothermic)

cations(g)

+

anions(g)

(exothermic)

(iii) =

ΔHθhyd(cations(aq)) + ΔHθhyd(anions(aq))

sum of the enthalpies of hydration of the cations and anions

 

(i) = (ii) + (iii), so in general :–

ΔHθsolution,298K(ionic compound) = –ΔHθLE(compound(s)) + ΔHθhyd(cations(g)) + ΔHθhyd(anions(g)) kJmol–1

From above the processes (i), (ii) and (iii) are now put together in a Hess's Law cycle for sodium chloride.

(i) ΔHθsolution,298K(sodium chloride) = +4 kJmol–1
(i) NaCl(s) + aq (c) doc b Na+(aq)  +  Cl–(aq)                

(ii) = –ΔHθLE(NaCl(s))

– lattice enthalpy of sodium chloride

(c) doc b

(iii) = ΔHθhyd(Na+(aq)) + ΔHθhyd(Cl–(aq))

sum of the enthalpies of hydration of the sodium ion and chloride ion

Na+(g)  +  Cl–(g)                     
(i) = (ii) + (iii)

ΔHθsolution,298K(sodium chloride) = –ΔHθLE(NaCl(s)) + ΔHθhyd(Na+(g)) + ΔHθhyd(Cl–(g)) kJmol–1

+4 = –(–774) + (–406) + (–364) = +774 –406 –364

Example 2 Dissolving the alkali potassium hydroxide

(i) ΔHθsolution,298K(potassium hydroxide) = ? kJmol–1
(i) KOH(s) + aq (c) doc b K+(aq)  +  OH–(aq)                               

(ii) = –ΔHθLE(KOH(s))

– lattice enthalpy of sodium chloride

(c) doc b

(iii) = ΔHθhyd(K+(aq)) + ΔHθhyd(OH–(aq))

sum of the enthalpies of hydration of the sodium ion and chloride ion

K+(g)  +  OH–(g)                     
(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 (c) doc b Al3+(aq)  +  3F–(aq)                               

(ii) = –ΔHθLE(AlF3(s))

– lattice enthalpy of sodium chloride

(c) doc b

(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)                     
(i) = (ii) + (iii)

ΔHθsolution,298K(aluminiumfluoride) = –ΔHθLE(AlF3(s)) + ΔHθhyd(Al3+(g)) + 3 x ΔHθhyd(F–(g)) kJmol–1


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 quite high (quite 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.
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 and in the case of a slightly exothermic enthalpy of solution the entropy change becomes most important (see below).
  • 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!

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 most 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 most likely to be insoluble. Even the solubility of salts in organic polar molecules like ethanol is usually quite low.

However, if the solute is NOT ionic then there is no lattice enthalpy to overcome so small covalent molecules like iodine will dissolve in hexane or ethanol.

–


See Part 2.2 for Born–Haber Cycle Calculations


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