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?

Yes, there is more to consider than just the dissolving process when determining the sign of the enthalpy change (delta H). In the case of NaNO3 dissolving in water, even though delta G is negative and delta S is positive, the positive delta H indicates that the dissolving process is endothermic. This means that energy is absorbed from the surroundings during the dissolution, resulting in a decrease in the temperature of the system.

On the other hand, when NaCH3COO dissolves in water, the negative delta H indicates that the process is exothermic. In this case, energy is released into the surroundings during dissolution, leading to an increase in temperature.

The difference in the enthalpy changes between these two processes can be attributed to the intermolecular forces involved. NaNO3 has stronger ion-dipole interactions with water molecules, which require more energy to break, resulting in an endothermic process. However, NaCH3COO has weaker ion-dipole interactions, leading to a release of energy and an exothermic process.

So, in addition to delta G and delta S, the sign of delta H is determined by the strength of the intermolecular forces involved in the dissolving process. It is crucial to consider these factors to fully understand why one process can be endothermic while the other is exothermic, despite both having negative delta G and positive delta S.

Yes, there are additional factors to consider when determining whether a process is endothermic or exothermic apart from the dissolving process itself. While the sign of ΔG and ΔS do provide information about the spontaneity and disorder of a process, they do not directly indicate whether a process is endothermic or exothermic.

The enthalpy change (ΔH) of a process represents the amount of heat absorbed or released during the process. A positive ΔH indicates an endothermic process (heat being absorbed), whereas a negative ΔH indicates an exothermic process (heat being released).

In the case you mentioned, even though both processes have a negative ΔG (spontaneous) and a positive ΔS (increased disorder), the difference in the enthalpy changes results in one process being endothermic and the other being exothermic.

When NaNO3 dissolves in water, it absorbs heat from the surroundings to proceed. This means ΔH is positive (endothermic), even though the overall process is favored due to the negative ΔG and positive ΔS.

On the other hand, when NaCH3COO dissolves in water, it releases heat into the surroundings. Therefore, ΔH is negative (exothermic), and again, the overall process is favored due to the negative ΔG and positive ΔS.

So, in conclusion, the thermodynamic parameters ΔG and ΔS provide information about spontaneity and disorder, while ΔH determines whether a process is endothermic or exothermic.

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.