# Are there endothermic dissolution reactions? [closed]

Are there reactions like

$X(s) + n Y(l) \to X \centerdot Y_n(l)$

that are endothermic? What are X and Y then?

The material X should also be soluble to material Y.

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Note that we are currently discussing on the meta.chemistry.SE site how fairly basic questions should be handled… I suppose this question is an interesting example, and would welcome other to chime in below and give their opinion. –  F'x May 9 '12 at 21:21
I'll start with my own opinion: I think this question is easily answerable by Googling (which I did to copy-paste the text below) or opening a textbook, and as such should be closed. –  F'x May 9 '12 at 21:22
In my opinion, this is not clear as it has two parts (that are not in the same page in wikipedia): Substances should be soluble and the process should be endothermic. One crucial part of the answer is that there should be enough heat (solubility is more dependent of heat in endothermic reactions than in exothermic reactions, right?). Noted, that this was a simple question that after seeing the answer I should have known. –  Juha May 9 '12 at 21:29

## closed as off topic by F'x♦, Andrew, ManishEarth♦May 29 '12 at 12:37

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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.

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

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How can the reaction happen if it is endothermic? –  Juha May 9 '12 at 21:13
@Juha: Take energy from the surroundings, of course. Lots of reactions are endothermic. Whether they happen or not is determined by $G$, not $H$ –  ManishEarth May 23 '12 at 4:29