Why are IUPAC's definitions of exo- and endothermic disconnected from the direction of heat flow?

I'm wondering if anyone can provide a rationale for IUPAC's definitions of exothermic and endothermic, since they disconnect these terms from the direction of heat flow.


Specifically, here are IUPAC's *Gold Book* definitions:



> 
> **exothermic reaction** (<https://doi.org/10.1351/goldbook.E02269>, Last updated: February 24, 2014 (v. 2.3.3)):  
> 
>  A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.
> 
> 
> **endothermic reaction** (<https://doi.org/10.1351/goldbook.E02095>, Last updated: February 24, 2014 (v. 2.3.3)):  
> 
>  A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is positive.
> 
> 
> 


The superscript "$\circ$" indicates standard state, which is $\pu{1 bar}$ for a gas phase reaction (<https://doi.org/10.1351/goldbook.S05927> and <https://doi.org/10.1351/goldbook.S05921>, both last updated: February 24, 2014 (v. 2.3.3)) 


Suppose we have a gas-phase reaction for which $\Delta H^\circ < 0$. This is defined by IUPAC as an exothermic reaction. Correspondingly, if the reaction is carried out isothermally at standard state, $q\_p = \Delta H^\circ < 0$, indicating that there will be heat flow out of the reaction mixture. This comports with the common understanding of "exothermic". 


**But:** Suppose we study the same reaction at the same temperature, but a higher pressure. If $|\Delta H^\circ|$ is sufficiently small, and $\left(\partial ΔH\over\partial p\right)\_T$ is sufficiently large and positive then, with a sufficient increase in $p$, we could readily reach a point at which $\Delta H > 0$, and thus where (again, assuming isothermal and isobaric conditions) $q\_p > 0$, i.e., a point at which the reaction absorbs, rather than releases, thermal energy. Yet (according to IUPAC) we must still call that reaction "exothermic", even under those conditions, because $\Delta H^\circ < 0$. 


*I.e., according to IUPAC, if a reaction is exothermic at standard state, we must always call it exothermic, regardless of the pressure at which we're studying it, and the direction of heat flow at that pressure.* 


To my mind, this goes against both the common usage/understanding of the terms exothermic and endothermic within the field of chemistry, and also against the inherent meaning of the words themselves. For instance, if a group is studying a reaction in Jupiter's atmosphere that has $\Delta H\_p > 0$ under those conditions, but $\Delta H^\circ < 0$, they're supposed to call it "exothermic" when they discuss it in a paper. This is very confusing.


This could be resolved if IUPAC replaced "overall standard enthalpy change $\Delta H^\circ$" with "overall isobaric enthalpy change $\Delta H\_p$" in its respective definitions. Given this, can anyone explain why IUPAC chose to tie the definitions of exothermic and endothermic to $\Delta H^\circ$ rather than $\Delta H\_p$?


N.B. #1: There's also the broader question of whether, to match standard usage in the field, the definitions of exo/endothermic should be untethered from $\Delta H$ generally and, correspondingly, from isobaric processes. I.e., perhaps the definition needs to be made less restrictive still, accounting for both constant-*p* and constant-*V* processes. I don't do any thermochemistry, but don't researchers using bomb calorimeters (constant *V* rather than constant *p*) describe reactions that evolve thermal energy under those conditions as exothermic? If so, the criterion used there to determine if a reaction is exo- or endothermic is the sign of $\Delta U\_V$ rather than $\Delta H\_p$. 


Alternately, we could abandon pressure-volume restrictions altogether, and tie the definition directly to heat flow instead of a state function: 'An exothermic reaction is one that, if carried out isothermally, has $q < 0$. This means that: (1) If it is carried out at constant *p*, $\Delta H < 0$; (2) If it is carried out at constant *V*, $\Delta U < 0$; (3) if it is carried out adiabatically ($q = 0$), $\Delta T > 0$.' But I don't think IUPAC would wish to go that far.


N.B. #2: There is an alternate interpretation of IUPAC's definition, which is that endothermic or exothermic are only defined at standard state—i.e., that if a reaction takes place outside of standard state, one cannot use the terms exothermic or endothermic at all. But this also goes against common practice.


