# What causes the entropy to change when a solute dissolves in a solvent?

Jun 18, 2017

The entropy increases for formation of an ideal solution due to the dispersion of the solute particles' energy amongst the solvent particles, assuming no intermolecular interactions form (i.e. that the change in enthalpy of mixing is neither endothermic nor exothermic).

This is due to the definition of entropy being the amount of energy dispersal, and that energy was dispersed after mixing.

When intermolecular interactions are not formed between solute and solvent particles after mixing (i.e. the formation is neither endothermic nor exothermic), all of the solute-solute interaction energy that was released from the solute particles is dispersed throughout the resultant solution, increasing the entropy after mixing. For ideal two-component solutions, the change in entropy of mixing is given by:

$\Delta {S}_{m i x}^{\text{id}} = - R \left[{n}_{i} \ln {\chi}_{i} + {n}_{j} \ln {\chi}_{j}\right]$

where:

• $\Delta {S}_{m i x}^{\text{id}}$ is the change in entropy due to mixing. The initial state is the unmixed solute and solvent, and the final state is the mixed solute and solvent.
• $n$ is the mols of solute $i$ or solvent $j$.
• ${\chi}_{i} = \frac{{n}_{i}}{{n}_{i} + {n}_{j}}$ is the mol fraction of solute $i$ in a two-component solution.
• $R = \text{8.314472 J/mol"cdot"K}$ is the universal gas constant.

Since mol fractions are always $\le 1$, $\ln \chi \le 0$ and thus the change in entropy of mixing for ideal solutions is always positive.