If the sign for delta G is negative (spontaneous process) and sign for delta S is positive (more disorder) for both dissolving processes, how could one be endothermic (positive delta H- NaNO3 dissolves in water) and one be exothermic (negative delta H- NaCH3COO dissolves in water)? Is there more to consider than just the dissolving process?

Question

DrBob222 · Answer

I don't think so. If dS is + then the term -TdS will always be -. So dG will be - if dH is not too + to more than balance out the -TdS term.