few compounds have a decreased solubility with greater temp. how would H, S and G differ from an endothermic?

Question

how would it specifically differ from the dissolution of potassium nitrate? how would enthalpy, entropy and gibbs free energy be different?

Answer 1

We'll take this one at a time.

The free energy is easy: a reaction will only proceed naturally if ΔG is negative. Even though a solute may have LESS solubility at higher temperatures, it DOES dissolve. So the free energy must still be negative, albeit less so.

The one thing temperature has an effect on is the effect of entropy (ΔG = ΔH-TΔS, after all). Normally when something dissolves it becomes more soluble at higher temperatures for this reason - the entropy increases, this has an inverse effect on free energy, so it's more prone to go! If you have LESS solubility with higher temperature, the reverse must be happening. Instead of ΔS being positive, it must be negative, or in other words the solute is more disordered on its own than when dissolved. This is the primary difference between the two solutes!

Now if you add the two things we've got so far, it will tell you what the ΔH is. ΔG must be negative. The entropy is making the reaction NOT go, so -TΔS is positive overall. So the only way this reaction works at all is if ΔH is negative. This makes sense if you think about it, too - if you dump heat into a substance, you'll probably increase its temperature, as well as make it harder to be exothermic. So your solute with lower solubility at higher temperature probably has a greater ΔH than the other one.

Hope that helps!