Can endergonic reactions be exothermic?

Pretty much the title. I've heard it said that no (see, e.g [here](https://www.quora.com/Is-an-exergonic-chemical-reaction-always-exothermic), answer by Lukas Schaedler). To quote his argumentation:



> 
> No. Exergonic reactions may be both exothermic or endothermic. Endergonic reactions are endothermic only.
> 
> 
> Endergonic and exergonic relate to changes in free energy (delta G), while endothermic and exothermic are related to changes in enthalpy (delta H).
> 
> 
> Gibb’s equation written in form of free energy (G), enthalpy (H) and entropy (S) changes:
> 
> 
> delta G = delta H - T . delta S
> 
> 
> Therefore, it’s possible for an exergonic reaction (delta G < 0; decrease in free energy) to be both exothermic (delta H < 0; heat is released) or endothermic (delta H > 0; heat is consumed). The second scenario may happen if :
> 
> 
> T . delta S > delta H
> 
> 
> On the other hand, endergonic reactions are endothermic only; delta H needs to be positive and greater than T . delta S in order to delta G > 0
> 
> 
> 


But it's not really clear to me why. In particular, in the example given, why is $\Delta S \ge 0$ (which in my understanding would be necessary to force $\Delta G = \Delta H - T\Delta S$ to be negative)?




The person who posted that answer on Quora was probably confused about the significance of $\Delta S$ as it relates to spontaneity. For isolated systems in general the sign of $\Delta S$ (the change in the entropy of the *system*) is the relevant criterion determining whether a process is spontaneous, since the entropy of the surroundings remains constant, so changes in the entropy of the universe are limited to the system.


$\Delta G$ is of course the relevant criterion for constant temperature and pressure processes. When the change is negative (exergonic) the process is spontaneous. This places no constraints on $\Delta H$ other than that it should satisfy


$$\Delta H \le T\Delta S$$


which can be interpreted as meaning that changes in the entropy of the surroundings must more than compensate for any potential decrease in the entropy of the system: if the entropy of the system decreases (is negative) then the entropy change of the surroundings ($-\Delta H/T$) must be positive and greater than this in magnitude:


$$\Delta S\_\textrm{surr} \ge -\Delta S$$


Freezing of any pure substance above its melting point is an example of a process that is not spontaneous but exothermic. By extension, the reverse is also true: melting of a substance above its freezing point is exergonic (spontaneous) yet requires an input of heat (endothermic).


Note also that $\Delta G^\circ$ determines the position of equilibrium since $\Delta G^\circ = -RT \log K\_\textrm{eq}$, but does not suffice to say whether the reaction is spontaneous. For that you need to compute $\Delta G = \Delta G^\circ + RT \log Q$, that is you need information about the starting concentrations.


 2