The dissolution of ammonium chloride is used to cool a container of water placed in the solution. It's an endothermic process. What absorbs the heat and what loses it?


When discussing this question, one needs to think about all the processes involved (or at least those that account for most of what happens). In general, any process will occur spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.

$$\Delta G = \Delta H - T \Delta S\tag{1}$$

There is an enthalpic term related to the heat a reaction produces or requires and an entropic term that is temperature-dependent. We cannot say a lot about the entropic term a priori, but we can about the enthalpic term — which is luckily what the question is about. For that, we first need to write a reaction equation (2):

$$\ce{NH4Cl (s) + n H2O <=>> NH4+ (aq) + Cl- (aq) + m H2O}\tag{2}$$

Dissolving of the ammonium chloride can also be thought as two separate processes, see equation $(2')$.

$$\ce{NH4Cl (s) -> [NH4+ (g)] + [Cl- (g)] -> NH4+ (aq) + Cl- (aq)}\tag{2'}$$

Meaning that first we break up the ammonium chloride crystal structure and second we dissolve the bare ions. If we want to write that in single enthalpy terms, we can do that as shown in equation (3).

$$\Delta H_\mathrm{tot} = -\Delta H_\mathrm{lattice}(\ce{NH4Cl}) + \Delta H_\mathrm{solv}(\ce{NH4+}) + \Delta H_\mathrm{solv}(\ce{Cl-})\tag{3}$$

All of these values can be looked up. I have given those as calculated by Jenkins and Morris in table 1.[1]

$$\textbf{Table 1:}\text{ Values of enthalpies used in this answer as}\\\text{quoted from Jenkins and Morris (reference 1).}\\\begin{array}{cccc}\hline \text{compound} & \Delta H_\mathrm{lattice} [\mathrm{kJ/mol}] & \Delta H_\mathrm{solv} [\mathrm{kJ/mol}] & \Delta H_\mathrm{tot} [\mathrm{kJ/mol}] \\\hline \ce{NH4Cl} & -709.1 & - 694.7 & 14.4 \\ \hline\end{array}$$

So we need energy (a lot of it!) to break up the crystal lattice of $\ce{NH4Cl}$. We then re-gain energy by creating the hydrated, i.e. dissolved, ions in solution. If energy is required, it is typically (excluding photochemical reactions — not the case here) heat energy which is simply drawn from the surroundings. If energy is released, it is typically (same caveat) released as heat into the surroundings.

Thus, the answer to your question is:

  • The heat is absorbed by the solid ammonium chloride. It is used to break up the salt crystal according to equation $(2'.1)$. Part of it is re-released (unnoticed) by forming solvent–ion hydrogen bonds (the solvation mechanism of $\ce{NH4Cl}$ — equation $(2'.2)$).

  • This heat is drained from the surroundings, which in this case is primarily the water in which you wanted to dissolve the ammonium chloride. (Of course, the solution will then further exchange heat with whatever is around it, meaning that either the air around a bottle/flask or an outer flask as in your described experimental setup will lose heat at the benefit of the $\ce{NH4Cl}$-solution.)


[1]: H. D. B. Jenkins, D. F. C. Morris, Mol. Phys. 1976, 32, 231. DOI: 10.1080/00268977600101741.


To keep it simple, the "heat" is lost by the water and the "ammonium chloride" has a net gain.

The ammonium chloride starts off as a solid crystal. The positive aluminum ions and the negative chloride ions attract, so it takes energy (as "heat") to break the crystal apart. The solvation of the anions and cations releases energy, but in this case it isn't enough to offset the energy spent breaking the crystal apart. So overall solution cools.

The thing to remember here is that for salts the "molecules" don't exist per sey in solution. Thus we could make a solution which is 1.00 molar in potassium chloride and 1.00 molar in sodium nitrate and we'd get the same solution as 1.00 molar potassium nitrate and 1.00 molar sodium chloride.


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