When stating this question, one requirements to think around all the processes affiliated (or at least those that account for most of what happens). In general, any procedure will take place spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.
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$$\Delta G = \Delta H - T \Delta S\tag1$$
There is an enthalpic term related to the warmth a reaction produces or requires and also an entropic term the is temperature-dependent. We cannot to speak a lot around the entropic term a priori, yet we can around the enthalpic ax — i m sorry is luckily what the question is about. For that, we an initial need to compose a reaction equation (2):
$$\ceNH4Cl (s) + n H2O > NH4+ (aq) + Cl- (aq) + m H2O\tag2$$
Dissolving of the ammonium chloride can likewise be assumed as two different processes, view equation $(2")$.
$$\ceNH4Cl (s) ->
Meaning that very first we break up the ammonium chloride crystal framework and 2nd we dissolve the ceiling ions. If we want to create that in single enthalpy terms, we can do the as displayed in equation (3).
$$\Delta H_\mathrmtot = -\Delta H_\mathrmlattice(\ceNH4Cl) + \Delta H_\mathrmsolv(\ceNH4+) + \Delta H_\mathrmsolv(\ceCl-)\tag3$$
All of this values can be looked up. I have given those together calculated by Jenkins and Morris in table 1.<1>
$$\textbfTable 1:\text values of enthalpies supplied in this prize as\\\textquoted indigenous Jenkins and also Morris (reference 1).\\\beginarraycccc\hline\textcompound & \Delta H_\mathrmlattice <\mathrmkJ/mol> & \Delta H_\mathrmsolv <\mathrmkJ/mol> & \Delta H_\mathrmtot <\mathrmkJ/mol> \\\hline\ceNH4Cl & -709.1 & - 694.7 & 14.4 \\ \hline\endarray$$
So us need energy (a lot of it!) to rest up the crystal lattice the $\ceNH4Cl$. 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 — no the case here) heat power which is simply attracted from the surroundings. If energy is released, that is typically (same caveat) exit as heat into the surroundings.
Thus, the answer come your inquiry is:
The warmth is absorbed by the heavy ammonium chloride. The 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 binding (the solvation device of $\ceNH4Cl$ — equation $(2".2)$).
This heat is drained from the surroundings, which in this case is generally the water in i beg your pardon you want to dissolve the ammonium chloride. (Of course, the equipment will then additional exchange warm with everything is around it, meaning that one of two people the air about a bottle/flask or an outer flask together in your described experimental setup will lose heat at the benefit of the $\ceNH4Cl$-solution.)
<1>: H. D. B. Jenkins, D. F. C. Morris, Mol. Phys.
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1976, 32, 231. DOI: 10.1080/00268977600101741.