# Question #87739

##### 1 Answer
Jul 14, 2017

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:

$\boldsymbol{\Delta {S}_{m i x} = {\sum}_{i = 1}^{2} {n}_{i} \left({\overline{S}}_{i} - {\overline{S}}_{i}^{\text{*}}\right) = - n R \left[\ln {\chi}_{1} + \ln {\chi}_{2}\right]}$

where:

• ${\overline{S}}_{i}$ is the molar entropy of substance $i$ in the mixture. $\text{*}$ 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 $0$ and $1$ by virtue of being fractions out of $1$. As a result, $\ln {\chi}_{i} < 0$ for two substances (like salt and water) combined together.

Therefore...

$\textcolor{b l u e}{\underline{\Delta {S}_{m i x}}} = - n R \left(\ln {\chi}_{1} + \ln {\chi}_{2}\right)$

$= - \left(+\right) \left(+\right) \left[\left(-\right) + \left(-\right)\right] \underline{\textcolor{b l u e}{> 0}}$

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.