The hydration energy of NaCl is −774.3 kJ. Then why are we not able to harness that energy and run machines just by mixing salt in water?

Where does all the energy go which is released when we mix salt with water?

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    $\begingroup$ The dissolution of NaCl in water is slightly endothermic, so the one number you have does not tell the whole story. $\endgroup$
    – Ed V
    Feb 15, 2020 at 14:54
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    $\begingroup$ Do you actually have any salt at home? Try to mix it with water. Go on, do it. Chemistry is an experimental science, after all. Got any energy? $\endgroup$ Feb 15, 2020 at 16:03
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    $\begingroup$ @IvanNeretin With all due respect, this isn't helpful. It sounds like you're giving the questioner a hard time for asking the question—something I've noticed you often do to first-time questioners. This seems like a violation of the SOC to be nice to new contributors. In addition, your comment implies one can't get energy from dissolving salt in water, which is incorrect. $\endgroup$
    – theorist
    Feb 15, 2020 at 18:05
  • $\begingroup$ That energy is in kJ just from one mole NaCl. And where does all the energy go? $\endgroup$ Feb 15, 2020 at 18:25
  • $\begingroup$ Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! $\endgroup$
    – Ed V
    May 26, 2020 at 1:40

2 Answers 2


EDIT: I just did the experiment that Ivan Neretin suggested. I used 250 mL of water in a 400 mL PTFE beaker. I measured the water temperature with a Hg thermometer and got 20.5 degrees Celcius. I then added 14.51 g NaCl, stirred gently to full dissolution, and the new measured temperature was 19.5 degrees Celcius. So the salt water was about 1 M and the temperature dropped by 1 degree.

From the wiki article on "Enthalpy change of solution", we have the following quote:

Wiki quote

The first two steps are endothermic and only the third is exothermic. According to the OP, that last step has hydration (i.e., aqueous solvation) enthalpy of -774.3 kJ/mol. But the sum of the first two steps, via Born-Haber cycle, is +778.2 kJ/mol, so the sum is +3.9 kJ/mol. Thus, we see that dissolving salt in water is slightly endothermic, as shown in the table from the wiki article:

Some solution enthalpies

So the hydration enthalpy is not for the entire dissolution process, i.e., solid NaCl dissolving in water. It is only for the last of the three steps.

An alternative to the three conceptual steps is to combine the second and third steps. As before, the first step simply involves putting in sufficient energy to convert the salt crystal to ions that are far enough apart so that coulombic interactions can be ignored. Then the combined second and third steps conceptually involve putting all those ions into water. This scheme is illustrated, for KF, in the following figure (© Brooks/Cole, Cengage Learning):

KF solution enthalpy

From the figure, the enthalpy of solution is exothermic, but only -16 kJ/mol, while the enthalpy of hydration is -837 kJ/mol.

Bottom line: the OP's confusion was in assuming enthalpy of hydration was equal to enthalpy of solution.


The amount energy available to do work is the free energy (not the enthalpy). If you dissolve NaCl in water at 25 °C until you reach saturation, the Gibbs free energy change is about −8 kJ/mol. You can indeed harness this energy by using it to generate osmotic pressure, which can in turn be used to, say, raise a weight in the surroundings. [Strictly speaking, since this is not a constant-pressure process, the Gibbs free energy doesn't exactly give the energy available to do work; but since there is little volume change, the discrepancy is small.]


The osmotic pressure apparatus required to harness energy in significant quantities would be complex, and the amount of energy produced is quite small (relative to that obtained, say, from burning natural gas). World salt production is 259, 000,0 00 metric tons/year. Even at 100 % energy conversion of the dissolution of all of that salt, the plant would only generate 1 MW! That's likely significantly less than the energy cost of doubling current world salt production (even if that were possible) (needed to feed the plant at that level), as well as other energy costs associated with plant operation, maintenance, and construction.

In sum, you'd be losing energy (and a lot of money).

Finally, generating energy from turning fresh water into salt water is the opposite of what humanity is currently working to do: instead, we are using energy to turn salt water into fresh, with desalination plants.

  • $\begingroup$ Do you have any estimates for the energy input required for desalination? I know it is something of an apples and oranges comparison, given non-standard states, but it would be interesting to have a ballpark estimate. $\endgroup$
    – Ed V
    Feb 15, 2020 at 18:28
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    $\begingroup$ @EdV Accoring to en.wikipedia.org/wiki/Desalination#Energy_consumption, the most efficient desalination plants use 3 kWh/m^3 to remove salt from seawater. However, with reverse osmosis, it is possible to achieve desalination of seawater using < 2 kWh/m^3. Given that the molarity of salt in seawater is ~0.6, that works out to 18 kJ/mol and < 12 kJ/mol, respectively (where "mol" refers to moles of NaCl removed). $\endgroup$
    – theorist
    Feb 15, 2020 at 19:05

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