Feel like you can explain all of these phenomena with Gibbs free energy. For an exothermic interaction (-dH) if the entropic interaction is unfavorable (+dS) at high temperatures the entropic interaction component will make your dissolution unfavorable (+dG). Assuming solvating Calcium acetate not favorable entropically. However in the case of NaOH, this is both exothermic (-dH) and entropically favorable (-dS) so the increased temp will only make the interaction (dissolving) more favorable (-dG).

Adding heat to an exothermic reaction to slow it down - there's probably some people who have been involved with horrible runaway reactions that would beg to differ.

Oxidation of a fuel (Fire) certainly provides evidence that the slowdown of exothermic reactions is not what he thinks it is, otherwise it would quickly extinguish itself.

Acctully, I think he is right about it... Not about all (especially when he tries to talk about kinetics and the phenomenon is thermodynamic ). But he's right about increasing temperature in exothermic process, change the equilibrium, and it's more favorable to the products, because it changes 𝐾 constant. ΔG = ΔG° +RT•Ln(Q) At equilibrium 0 = ΔG° +RT•Ln(𝐾) Ln(𝐾) = -ΔG°/RT Ln(𝐾) = -(ΔH° -T•ΔS°)/RT Ln(𝐾) = -(ΔH°/RT) + ΔS° Considering an exothermic process in two different temperatures T₁ and T₂ and supposing T₂ >T₁ Ln(𝐾₁) = ΔH°/RT₁ + ΔS° Ln(𝐾₂) = ΔH°/RT₂ + ΔS° Then Ln( 𝐾₁) > Ln( 𝐾₂) for the same exothermic process, therefore it is better decrease temperature to make more products, in that case, solvateded ions.

Exactly, he seems to have dismissed Chatelier's principle; exothermic reactions (like calcium acetate's dissolution) are disfavored on higher temperatures since the heat absorption from the reverse reaction becomes more entropically favorable (dQ = T*dS) and the equilibrium will shift "left" to maximize entropy in the system. James did get the right idea.

Technically, he did account for entropy. In a simple 1-reaction no other phase change situation, Le Chatelier's principle applied to heat energy has entropy baked-in to the analysis. The argument goes as follows, using the equation dG (Gibbs Free Energy)=dH(enthalpy)-T(temp in K, effectively always positive) * dS (entropy change): 1) The Gibb's Free energy (accounting for current concentrations) at saturation will always be exactly zero by definition; otherwise, either more would dissolve (dG<0) or it would precipitate (dG>0) 2) While we usually consider the dissolution of solids in liquids entropically favorable (dS>0), that's not true at high concentrations of solute, so mixing can actually be entropically unfavourable (dS<0). 3) While we usually consider the dissolution of gases in liquids entropically unfavorable (dS<0), that's not true at low concentrations of solute, so mixing can actually be entropically favorable (dS>0). 4) Therefore, any dS is possible. 5) For an exothermic dissolution (almost all gases and some solids), since dH<0, if dG=0, then dS<0; that is, the saturation point will be such that dissolution is entropically unfavorable (even for solids!). Increasing T then increases dG, driving the equilibrium in reverse, causing lower saturation point and precipitating the solid/releasing the gas. 6) For an endothermic dissolution (some solids and rare cases of certian gases in certian liquids), since dH>0, if dG=0 then dS>0; that is, the saturation point will be where dissolution is entropically favorable (even for the rare gas case!). Increasing T then decreases dG, driving the equilibrium forwards, causing lower saturation point and dissolving more solute. In conclusion, the observations match for both, making the two concepts equivalent for this description! aka, it's different ways of explaining what is fundamentally the same thing.