Is it possible for a dissolution reaction of the form

$$\ce{X(s) ->[solvent] X (solv)}$$


$$\ce{M_mX_x (s) ->[solvent] m M^x+ (solv) + x X^m- (solv)}$$

to be endothermic, i.e. $\Delta_\mathrm r H > 0$?


1 Answer 1


You will find the answer in probably any textbook on chemical thermodynamics, or in a thermochemistry chapter in most physical chemistry textbooks, or in Wikipedia's enthalpy of dissolution page:

Dissolution can be viewed as occurring in three steps:

  1. Breaking solute-solute attractions (endothermic), see for instance lattice energy in salts.
  2. Breaking solvent-solvent attractions (endothermic), for instance that of hydrogen bonding
  3. Forming solvent-solute attractions (exothermic), in solvation.

The value of the enthalpy of solution is the sum of these individual steps. Dissolving ammonium nitrate in water is endothermic. The energy released by solvation of the ammonium ions and nitrate ions is less that the energy absorbed in breaking up the ammonium nitrate ionic lattice and the attractions between water molecules. Dissolving potassium hydroxide is exothermic, as more energy is released during solvation than is used in breaking up the solute and solvent.

So, it can have any sign. The linked page also gives data for common compounds in water, some of them endothermic, some of them exothermic. Ammonium nitrate, $\ce{NH4NO3}$, is a good example of a salt which dissolves in water in an endothermic reaction: if you dissolve some $\ce{NH4NO3}$ in water, the drop in the temperature will be noticeable.

Regarding why the substance is soluble why its dissolution is endothermic, you have to remember that the reaction takes place if $\Delta_\mathrm r G$ is favourable (i.e. negative), and $\Delta_\mathrm r G = \Delta_\mathrm r H - T \Delta_\mathrm r S$. Overall the free energy must be negative for dissolution to occur (on a thermodynamic basis; kinetics are another issue), not the enthalpy.

  • $\begingroup$ How can the reaction happen if it is endothermic? $\endgroup$
    – Juha
    May 9, 2012 at 21:13
  • 4
    $\begingroup$ @Juha: Take energy from the surroundings, of course. Lots of reactions are endothermic. Whether they happen or not is determined by $G$, not $H$ $\endgroup$ May 23, 2012 at 4:29

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