IB DP Chemistry: HL复习笔记15.1.4 Dissolution Energy Cycles

• In order to calculate either ΔH_sol^ꝋ , ΔH_lat^ꝋ or ΔH_hyd^ꝋ from given data we must apply Hess's Law

Energy cycle showing the application of Hess's Law to sodium chloride

• The energy cycle shows that there are two routes to go from the gaseous ions to the ions in an aqueous solution:
  • • Route 1: going from ionic solid → ions is the gaseous phase → ions in aqueous solution (this is the indirect route)
    • Route 2: going from ionic solid  →  ions in aqueous solution (this is the direct route)
• According to Hess’s law, the enthalpy change for both routes is the same, such that:

ΔH_sol^ꝋ = ΔH_lat^ꝋ + ΔH_hyd^ꝋ

• Each ion will have its own enthalpy change of hydration, ΔH_hyd^ꝋ, which will need to be taken into account during calculations
  • The hydration enthalpy is the sum of the hydration enthalpies of each ion
  • The total ΔH_hyd^ꝋ is found by adding the  ΔH_hyd^ꝋ values of both anions and cations together
  • If there is more than one cation or anion, such as in MgCl₂, then you must multiply by the appropriate coefficient for that ion

• This can also be represented as a Born-Haber cycle with the same direct and indirect route

Born-Haber cycle for sodium chloride

Answer:

Option 1 - Drawn as a Hess's Law cycle:

CaF₂ Hess’s Law Cycle

It is possible to complete this question by drawing a Born-Haber cycle, but examiners see mistakes more often on hydration and solution enthalpy questions when they are completed using a Born-Haber cycle.

Option 2 - Drawn as a Born-Haber cycle:

CaF₂ B-H cycle

Answer:

Drawn as a Hess's Law cycle:

• Hydration enthalpies are always negative values (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
• The polar water molecules will form ion-dipole bonds with the ions in solution causing the ions to become hydrated 
• The size of the hydration enthalpy is governed by the amount of attraction between the ions and the water molecules
• The smaller the ion, the stronger the attraction between the ions and the water molecules
  • As you go down a group, the ionic radius increases so attraction decreases and the the hydration enthalpy will become less exothermic
  • Overall, a smaller ion gives a more exothermic hydration enthalpy
• The more highly charged the ion; the stronger the attraction
  • The hydration enthalpies of 2+ ions in group 2 are much more exothermic than those of 1+ ions in group 1 as the attraction between the 2+ ions and the water molecules is stronger
  • Overall, a greater charge on the ion gives a more exothermic hydration enthalpy

Hydration enthalpies of group 1 and group 2 ions