We've already discussed in class a simple model for the formation of a solution composed of a solid solute and a liquid solvent. Each of the steps in this model involves an enthalpy and an entropy change. We know from the earlier introduction to thermodynamics in Unit 5 that the spontaneity of the dissolving process (i.e., will the solid dissolve or not) is based, in part, on the magnitudes and signs of these changes. Another factor is temperature:
DGo = DHo - TDSo
There is no question that the entropy of the final mixture is greater than that of the solid solute. Because the dissolved solute particles have access to many more modes of motion than they did when bound together in the solid state, there are many more pathways for energy dispersal. The entropy of the pure solvent state is also less than the final mixture based on a simple statistical argument.