# Question #87739

##### 1 Answer

It doesn't matter whether the temperature increases or decreases a little. The change in *entropy of mixing* for placing a solid into water is **positive** for ideal binary mixtures, period.

The enthalpy however, is endothermic with respect to the ions, sucking energy from the solvent to decrease its temperature at constant pressure...

The **entropy of mixing** for *ideal binary mixtures* (without the derivation) is given by:

#bb(DeltaS_(mix) = sum_(i=1)^(2) n_i(barS_i - barS_i^"*") = -nR[lnchi_1 + ln chi_2])# where:

#barS_i# is the molar entropy of substance#i# in the mixture.#"*"# indicates the unmixed substance(s).#n_i# is the mols of substance#i# in the mixture.#n# is the total mols of both substances in the mixture.#R# is the universal gas constant.#chi_i# is the mol fraction of substance#i# in the mixture.

Mol fractions are always between

Therefore...

#color(blue)(ul(DeltaS_(mix))) = -nR(ln chi_1 + ln chi_2)#

#= -(+)(+)[(-) + (-)] ul(color(blue)(> 0))#

And this should make physical sense. Dissolving things into a solvent disperses them throughout the solution.

That *spreads out the energy of the solute particles* more, which **increases** the entropy since the entropy is proportional to the amount of *energy dispersal*.