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When the process is exothermic (ΔHsystem​<0), and the entropy of the system increases (ΔSsystem​>0), the sign of ΔGsystem​ is negative at all temperatures. Thus, the process is always spontaneous.

It is easy to prove this using the formula of Gibbs free energy, but how can we prove it by logical reasoning?

I was thinking that in an exothermic process, the system loses heat. Shouldn't "the loss in heat" normally cause more ordering of the system and eventually a decrease in entropy? Might be a stupid question, but what am I missing here?

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    $\begingroup$ In an exothermic process, the system produces heat. Whether it loses that heat or keeps it to itself is an unrelated question. $\endgroup$ – Ivan Neretin Sep 8 at 8:12
  • $\begingroup$ Aha okay that clarified everything. Thank you $\endgroup$ – user208973 Sep 8 at 8:13
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It's a good idea to work out the assumptions implicit in your question. Since you are considering use of the Gibbs free energy as a criterion for spontaneity you are presumably concerned with a process carried out at constant T and P:

$$dG \leq 0 ~~~(\textrm{const. T, P)}$$

The second assumption is that $\Delta S$ and $\Delta H$, associated with the macroscopic process you are investigating, are independent of T and P over the range of interest. At constant T and P you are then allowed to write

$$ \Delta G = \Delta H - T \Delta S$$

for the entire range of interest, over which, again, $\Delta S$ and $\Delta H$ are assumed constant. Under these conditions, the above equation suffices to show that what you state in your first paragraph:

When the process is exothermic (ΔHsystem​<0), and the entropy of the system increases (ΔSsystem​>0), the sign of ΔGsystem​ is negative at all temperatures. Thus, the process is always spontaneous.

is true. The logical reasoning is as follows: a negative change in enthalpy corresponds at constant P and T to heat being released by the system. Of course this is what is understood as "exothermic". That heat increases the entropy of the surroundings. On the other hand, $\Delta S>0$ means the entropy of the system increases. Overall we then have that $$\Delta S_{sys}+\Delta S_{surr}>0$$ which guarantees spontaneity.

Regarding your second question, you should distinguish between changes in the entropy of the system at constant T and P, described by $\Delta S$, and changes in the entropy of the surroundings, described by (or proportional to) the heat emitted, or $\Delta H$. This is how you analyze, for instance, a reaction at constant T and P.

Consider instead a closed adiabatic system at constant P. Then

$$\left(\frac{\partial S}{\partial T}\right)_p = \frac{C_p}{T}$$

which means, since $C_p>0$ that, yes, as you state the entropy goes down if the temperature drops, all else being equal. But then T is not constant. When a reaction occurs or a piston moves, all else is not equal, and other sources of entropy production/depletion must be considered.

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Not all reaction is like this but, exothermic reactions are usually spontaneous because molecules always want to be more stable. So by reacting to another molecule and releasing energy, they can lower their own energy thus making them more stable and unreactive.

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  • $\begingroup$ My question was that why should being exothermic be accompanied by an increase in entropy and not a decrease in order for a rxn to be spontaneous (for all temperatures) $\endgroup$ – user208973 Sep 8 at 11:48

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