Enthalpies of solution and hydration
| English | Chinese | Pinyin |
|---|---|---|
| enthalpy of solution | 溶解焓 | róng jiě hán |
| enthalpy of hydration | 水合焓 | shuǐ hé hán |
| exothermic | 放热 | fàng rè |
| lattice energy | 晶格能 | jīng gé néng |
Dissolving, energetically
- Dissolving an ionic solid involves two enthalpy steps.
- Hydration releases energy; pulling the lattice apart costs it.
- A cycle links them to the enthalpy of solution 溶解焓.
Enthalpy of solution lab
deltaHsol = lattice + hydration terms
Change hydration strength and see solution enthalpy shift.
The enthalpy change when one mole of gaseous ions dissolves in water is the enthalpy of ______.
More negative for smaller, more highly charged ions.
Enthalpy of solution can be found from lattice energy and hydration enthalpies using Hess's law.
A Hess cycle links them.
The two changes
- enthalpy of hydration 水合焓 $\Delta H_{\text{hyd}}$: change when one mole of gaseous ions is surrounded by water (exothermic 放热).
- enthalpy of solution $\Delta H_{\text{sol}}$: change when one mole of solute fully dissolves.
Dissolving energy cycle links lattice energy 晶格能, hydration and solution
The enthalpy change of hydration is:
Water molecules are attracted to the ions, releasing energy, so hydration is exothermic.
The dissolving cycle
To dissolve, you first pull the lattice apart, then hydrate the ions:
- Like lattice energy, $\Delta H_{\text{hyd}}$ is more exothermic for ions of higher charge and smaller radius.
The enthalpy of solution is given by:
You reverse the lattice energy (to separate the ions) then add the hydration enthalpy.
The enthalpy of hydration is more exothermic for ions with:
Higher charge and smaller radius pull the water molecules in more strongly, releasing more energy.
Linking the enthalpies
- Enthalpy of solution = (−lattice energy) + enthalpies of hydration of the ions.
- A Hess cycle links the three quantities.
You've got it
- hydration $\Delta H_{\text{hyd}}$ (exothermic): gaseous ions surrounded by water
- solution $\Delta H_{\text{sol}}$: one mole of solute dissolves
- the cycle: $\Delta H_{\text{sol}} = -\Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$
- $\Delta H_{\text{hyd}}$ is more exothermic for higher charge and smaller radius