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I am used to thinking of exothermic reactions as being defined as having negative $ΔH.$ However, I also am used to hearing/thinking of exothermic reactions as "releasing heat".

Since change in enthalpy $∆H = Q + VΔP,$ it seems that enthalpy and heat flow $(Q)$ are not strictly the same thing. Is it not therefore possible to have a reaction which absorbs heat $(+Q),$ but has a decrease in pressure, such that the product $-VΔP > Q,$ thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by $Q,$ then an exothermic reaction with positive $∆H)?$

As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus $V∆P)$ with no heat flow or $PV$ work? $(Q = 0 = PΔV)$

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  • $\begingroup$ Exothermic reactions don't necessarily have to release heat, although they commonly do; they simply have to release excess energy. Heat doesn't always equal energy. $\endgroup$ – bio Jun 29 at 1:34
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"Since change in enthalpy ∆H = Q + VΔP, it seems that enthalpy and heat flow (Q) are not strictly the same thing."

First, note that, when you write "∆H = Q + VΔP," you are assuming the only type of work is PV-work. That's a reasonable assumption for these discussions. It's just important to make that assumption explicit.

Given this, the quoted statement is correct. Note, however, that many chemical reactions are carried out in an open container, such that the system stays at constant pressure (∆P = $0$) throughout the reaction. Thus, under this condition, ∆H = Q + VΔP = Q. I.e., the reason ∆H is a particularly useful thermodynamic function is that, under the common condition of ∆P = $0$, it corresponds directly to an important quantity: the amount of heat that flows into or out of the system.

"Is it not therefore possible to have a reaction which absorbs heat (+Q), but has a decrease in pressure, such that the product -VΔP > Q, thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by Q, then an exothermic reaction with positive ∆H)?"

Here it comes down to a definition of terms. The IUPAC Gold Book defines an "exothermic reaction" as "A reaction for which the overall standard enthalpy change ΔH⚬ is negative" ( https://goldbook.iupac.org/html/E/E02269.html ), where "standard" means P = 1 bar. Thus it doesn't cover how one would define exothermic or endothermic if the process is not at constant pressure.* Having said that you could, in principle, have a reaction that takes place, say, at constant volume, and that releases thermal energy (Q < $0$) but, because of a sufficiently large and positive VΔP term, has ΔH > $0$. But, in practice, $\mid{Q}\mid$ would have to be very small in order for the VΔP term to dominate.

"As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus $V∆P)$ with no heat flow or $PV$ work? (Q = $0$ = PΔV)"

Yes. Consider, for instance, an isolated system, which is a system in which no flow of heat or work is allowed, and that is closed (no flow of matter). The pressure could increase in an isolated system if, for instance, a chemical reaction is taking place in that system that evolves gas, and/or for which (were we to remove the adiabatic constraint) Q would be negative.

*Indeed, the IUPAC definition doesn't even seem to cover how one would define exothermic or endothermic if the reaction took place under constant P $\neq$ 1 bar, which seems quite limiting! Further, it only defines the term for a "reaction", which IUPAC seems to take to mean a "chemical reaction" (typing "reaction" into the search field redirects you to "chemical reaction"), thus excluding important processes like phase changes.

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  • $\begingroup$ I really have to wonder why IUPAC came up with that definition, rather than simply generalizing to endothermic meaning q>0, the latter afaik including the IUPAC definition. $\endgroup$ – Buck Thorn Jul 6 at 7:29
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    $\begingroup$ Yes, and a benefit of defining exo/endothermic in terms of the sign of Q would be that such a definition follows the natural meaning of the words: giving off or taking in thermal energy. The only thing I can think of is that the same chemical reaction has Q=$0$ if it's carried out adiabatically. Hence they may have decided it was safer to define it in terms of the change in a state function under certain conditions (constant P = 1 bar) rather than the integral of a path function (Q). Of course, they could have defined it using the sign of Q under isothermal conditions. $\endgroup$ – theorist Jul 6 at 15:25
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The heat of reaction $\Delta H^\circ$ is defined as the difference in enthalpy between the pure products and pure reactants, all at 1 bar pressure and 25 °C. If this $\Delta H^\circ$ is negative, the reaction is regarded as exothermic.

If we are dealing with all ideal gas species, where the enthalpy is independent of pressure and the heat of mixing is zero, and the reaction takes place in a process mixture maintained at constant pressure, the heat added to the reaction mixture to maintain constant temperature is equal to $\Delta H^\circ.$

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