# 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.

• Maybe this definition is supposed to provide a standard example rather than a strict definition, but it certainly is not worded that way. I find it odd. Commented Oct 6, 2019 at 11:56
• OMG. I stumbled upon this article: Understanding Endothermic and Exothermic Reactions The first paragraph makes me ask myself "where did she get a PhD from?" In any case and sliding away from the off-topic: The authors make the statement in the paper "Every effort has been made to include what appear to be the most commonly accepted definitions. [...] The reader is warned, however, that alternative definitions of some of the terms, [..], are frequently to be found in the scientific literature." (10.1351/pac199668010149) Commented Oct 6, 2019 at 12:50
• I think for the reasoning we would have to go back to the commission's reports, but I was unable to find it (probably not online, the earliest one in their database was from the end of 1996; even old iupac starts 1998). This is certainly an interesting query, but resources on the matter might be difficult to obtain. || On a technical note: the iupac tag was removed, see Do we need the [iupac] tag? Commented Oct 6, 2019 at 13:18
• @Martin-マーチン Yeah, I also searched through IUPAC's online resources prior to posting, and wasn't able to find an explanation. The only way to get an answer might be to contact IUPAC directly and get a referral to the appropriate person. I might do that at some point. But, since this is time-consuming, I was hoping someone on this site might have been involved with the relevant IUPAC commission and thus already have that institutional knowledge. Commented Oct 6, 2019 at 17:48
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Commented Oct 9, 2019 at 19:03