A colleague said we can’t dissolve a salt (whose solvation enthalpy is exothermic) faster if we increase the temperature (the solubility equilibrium product is not reached) because Le Chatelier‘s principle would favor the reactants.

For example, imagine dissolving NaOH(s) in distilled water, can’t this be accelerated by elevating the temperature?

I read a similar question on here, see: Factors that influence the kinetics of an irreversible exothermic reaction

How I see it is the following: As Le Chatelier‘s principle only applies to systems in equilibrium, and we only dissolve a salt (whose enthalpy happens to be exothermic), no equilibrium is established. The same goes for exothermic reactions in general. As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless whether it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in the Arrhenius equation).

It is important to note that this is about kinetics, that is, how quickly the NaOH dissolves, in a non-saturated solution, that is about 1M, not about how much NaOH a saturated solution can contain at a given temperature.

Is my reasoning correct?

  • 1
    $\begingroup$ Check this discussion: web.archive.org/web/20150228170949/http://… $\endgroup$
    – AChem
    Commented Dec 22, 2021 at 20:33
  • 2
    $\begingroup$ Does this answer your question? Why is heating increasing the solubility of sodium hydroxide? $\endgroup$
    – Mithoron
    Commented Dec 22, 2021 at 21:23
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    $\begingroup$ Not really, as the most upvoted answer talks about maximum solubility of a concentrated solution of NaOH, not about if a temperature increase dissolves a certain amount of NaOH (solid) faster $\endgroup$
    – Mäßige
    Commented Dec 22, 2021 at 21:34
  • $\begingroup$ A well-known and intuitive counter-example: burning (of everything). The reaction is exotermic and pretty much increases with temperature. This is how fires self-accelerate when not controlled. $\endgroup$
    – fraxinus
    Commented Dec 23, 2021 at 7:12
  • $\begingroup$ In this case, whether it is endothermic or exothermic depends on the concentration because a saturated solution is not an ideal solution, chemistry.stackexchange.com/a/4464/72973 $\endgroup$
    – Karsten
    Commented Dec 23, 2021 at 14:03

2 Answers 2


Typically, rates of uncatalyzed simple reactions increase with temperature. There are well-known examples where this is not the case (such as enzyme-catalyzed reactions where the enzyme denatures at high temperatures, or reactions with an intermediate that is at rapid equilibrium with the reactant in an exothermic step).

In this case, if the $\ce{NaOH}$ is soluble at both temperatures, it is likely that the increasing rate in the forward direction will make a bigger difference that the also increasing rate in the reverse direction.

As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless if it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in Arrhenius equation).

In this case, the reaction goes to completion rather than attaining equilibrium (if you stay below the solubility limit). Even for reactions that attain equilibrium, however, the rate with which they approach the equilibrium constant is proportional to the sum of the forward and reverse reaction (for a one-step reaction closes to equilibrium, it is possible to show this in a quick derivation).


Our current understanding of chemical kinetics is that all reactions with an activation energy are increased in rate by an increase in temperature. The question that is pertinent is "Which reaction is accelerated faster?". A temperature change at equilibrium influences the equilibrium and the overall composite rate. Whether the overall direction of a given reaction is accelerated by an increase in T is determined by the actual value of delta G; if negative it is accelerated as written.

The choice of NaOH as an illustration is a bit unfortunate. As anyone who has tried it knows adding water to solid NaOH results in a severe exothermic reaction even to a boiling solution. Yet NaOH is more soluble at higher T. This is contradictory to the idea that an exothermic reaction is lessened at a higher T. What happens is that the dissolution reaction becomes endothermic as the concentration increases; don't ask me how or why. The general answer: Displaced from equilibrium any reaction is accelerated by an increase in T provided there is an activation energy. At equilibrium the increase of both rates moves the equilibrium in the direction of the reaction with the higher activation energy. If the activation energy is vanishingly small the problem becomes removal of energy and the reaction doesn't exist.


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