# Can one predict the entropy change for aqueous phase reactions without calculation?

I have seen several sources that claim aqueous solutions always have more entropy than pure liquids or solids, however, some aqueous ions have negative standard entropy. This opens the possibility of solubility reactions with negative standard entropy change. For example, \begin{align} \ce{NaF (s) &<=> Na+ (aq) + F- (aq)}&&\Delta S^{\circ} =-6.26 \ \mathrm{J/(mol\cdot K)}\\ \ce{AlCl3 (s) &<=> Al^3+ (aq) + 3 Cl- (aq)}&&\Delta S^{\circ} =-261.5 \ \mathrm{J/(mol\cdot K)}\\ \ce{AlF3 (s) &<=> Al^3+ (aq) + 3 F- (aq)}&&\Delta S^{\circ} =-429.58 \ \mathrm{J/(mol\cdot K)} \end{align}

I have used thermodynamic data from this source.

So it seems to me that it is impossible to easily predict whether the entropy change for a reaction involving an aqueous phase is either positive or negative without calculation (unless a gas phase is involved, in which case the aqueous phase is moot)

Am I missing something, or are all the sources that claim an aqueous phase always has more entropy than a liquid phase oversimplifying the situation or just plain wrong?

I brought up the example of aluminum chloride to a chemistry instructor who then said that it is possible for molecular compounds to have negative entropy change when dissolved, but never for ionic compounds. Yet both sodium fluoride and aluminum fluoride are ionic, and seem to have a negative entropy change when dissolved.

I understand why the standard entropies are negative, and about solvation shells. What I'm asking is whether the change in entropy can be predicted at a glance instead of looking up values and calculating as I have done. Many textbooks claim all of these reactions should have a positive $\Delta S^{\circ}$ simply because "aqueous solutions have more entropy than solids or liquids." The instructor I talked to claimed that $\ce{AlCl3}$ will have negative $\Delta S^{\circ}$ because it is a molecular solid whereas $\ce{NaF}$ and $\ce{AlF3}$ should have positive $\Delta S^{\circ}$ because they are ionic solids. Are the textbooks and this instructor incorrect?

• BTW, aluminum chloride is also pretty much ionic when dissolved. – Ivan Neretin Apr 11 '16 at 22:47

Now, why are the calculations seemingly showing otherwise? The $\Delta S$ you have calculated above is the change in terms of hydration rather than dissolution. A negative $\Delta S_{hydration}$ is sensible because some ions do in fact impose a local order (relative to the proton) whereas others do not. However, if you want to look at the entropy of solution (which is then comparable to enthalpy of solution), you have to consider the case of infinite dilution.
Infinite dilution, by definition, is such that no more change in concentration is achieved despite adding more solvent. Assuming that the substance dissolves in the first place, at infinite dilution the individual solute molecules/ions are probably substantially far apart from each other such that they no longer interact with each other - they are totally free from each other but not the solvent. So relative to their 'bound state' of solid and liquid, dissolving the substance will then result in a $\Delta S$ that must be positive.