Free Energy of Dissolution
Introduced| English | Meaning |
|---|---|
| dissolution/ˌdɪsəˈluːʃn/ | dissolution |
Why some salts dissolve and others don't
- Table salt vanishes in water; chalk barely does.
- Both the heat and the disorder of dissolving matter.
- Together they decide whether dissolving is favourable.
- The same free-energy rule applies to dissolving.
Dissolving as a process
- Dissolution 溶解 has its own $\Delta G = \Delta H - T\Delta S$.
- If $\Delta G < 0$, dissolving is favourable.
- Both the heat and the entropy of dissolving count.
An instant cold pack: endothermic processes absorb heat from the surroundings
Dissolving energy cycle links lattice energy, hydration and solution
Dissolving is favourable when its $\Delta G$ is...
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$\Delta G < 0$ means dissolving is favourable.
Whether a salt dissolves favourably depends on...
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$\Delta G$ combines both $\Delta H$ and $\Delta S$.
Disorder favours dissolving
- Dissolving usually increases entropy, as the ordered solid spreads out.
- That positive $\Delta S$ pushes $\Delta G$ toward negative.
- So many salts dissolve even when it costs some heat.
Dissolving a solid usually increases the entropy.
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The ordered solid spreads out into solution.
An ordered solid spreading into solution represents an increase in ____.
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Spreading out raises the entropy.
The heat can go either way
- Some salts release heat as they dissolve; others absorb it.
- An endothermic dissolving can still be favourable if entropy wins.
- The balance of the two sets $\Delta G$.
An endothermic dissolving can still be favourable if...
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A large $T\Delta S$ can overcome a positive $\Delta H$.
A salt dissolves endothermically ($\Delta H > 0$) but with a large entropy gain. Can it still dissolve?
- The positive $T\Delta S$ can overcome the positive $\Delta H$.
- So $\Delta G < 0$ and it dissolves.
Will it dissolve?
Sort each case by whether dissolving is thermodynamically favourable.
Raising the temperature strengthens the $T\Delta S$ term, often helping a solid dissolve.
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A bigger $T\Delta S$ makes $\Delta G$ more negative.
An endothermic dissolving (which feels cold) can still be favourable if the entropy increase is large enough — do not assume endothermic means "will not dissolve." Both $\Delta H$ and $\Delta S$ of dissolving matter. And higher temperature strengthens the $T\Delta S$ term, often helping a solid dissolve.
Dissolution obeys $\Delta G = \Delta H - T\Delta S$ like any process. Dissolving usually raises entropy (a positive $\Delta S$ that pushes $\Delta G$ negative), so even an endothermic dissolving can be favourable if that entropy gain is large enough.