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I am confused as I can’t understand the difference between the thermodynamic terms q and ∆H.

In my book,“general chemistry” by Ebbing and Gammon,it is stated that at constant pressure Qp=∆U+P∆V=∆H.

How come can q equal ∆H? The only occasion at which q=∆H is when the moles of the limiting reagent in the thermodynamic equation is 1, such that q=∆H/n. What does constant pressure has to do with this equality? Because if we assume such equality of. Qp=∆U+P∆V=∆H at constant pressure, then it is also valid at constant volume when ∆V=0 such that Q=∆U=∆H.

Please help me understand this conundrum.

Many thanks in advance.

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  • $\begingroup$ You may need to dive a little more into the definition of enthalpy in other sources if your textbook does not cover it in detail, for instance look through other posts on this site, which probably answer your question. $\endgroup$
    – Buck Thorn
    Commented Jan 15, 2022 at 10:13
  • $\begingroup$ For instance: chemistry.stackexchange.com/questions/44129/… chemistry.stackexchange.com/questions/10187/… chemistry.stackexchange.com/questions/39988/… and many many more.... $\endgroup$
    – Buck Thorn
    Commented Jan 15, 2022 at 10:15
  • $\begingroup$ I think the core of the question is about the confusion between enthalpy vs molar enthalpy (extensive or intensive quantity), and the question about constant pressure is secondary. Textbooks are notoriously sloppy about this, switching from enthalpy to molar enthalpy without using a different symbol (or bold vs not bold like in upperlevel texts). Heat is always an extensive property. I think there is a good question here, so I voted to reopen. @BuckThorn $\endgroup$
    – Karsten
    Commented Jan 15, 2022 at 15:35
  • $\begingroup$ @KarstenTheis I agree about the confusion that caused the post and agree that this is a common issue for a lot of students starting thermo. Maybe you are right that the molar quantity issue is worth being addressed in particular, so reopen as I single-handedly closed. $\endgroup$
    – Buck Thorn
    Commented Jan 15, 2022 at 15:46
  • $\begingroup$ Related: chemistry.stackexchange.com/a/162243/72973 (which I just wrote to address both questions). $\endgroup$
    – Karsten
    Commented Jan 15, 2022 at 15:51

1 Answer 1

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In my book,“general chemistry” by Ebbing and Gammon,it is stated that at constant pressure Qp=∆U+P∆V=∆H.

This is correct in the absence of electrical and mechanical work (e.g. the equality breaks down for a battery or a fuel cell doing electrical work, or shaft work heating up the reaction mixture).

How come can q equal ∆H?

You can express the change in enthalpy as an extensive quantity (changes with the size of the reaction) or, in the case of molar enthalpy, an intensive quantity (does not depend on the amounts reacting). The equation above refers to enthalpy, the extensive quantity (with units of e.g. kJ). On the other hand, enthalpy of formation usually refers to the intensive quantity and should be called molar enthalpy of formation (with units of e.g. kJ/mol).

For the enthalpy of reaction, things are a bit fuzzy in introductory textbooks, either interpreting it as extensive or intensive. If it is used as intensive quantity, a change in the coefficients of the reaction (i.e. doubling or halving all coefficients) will change the value of the molar enthalpy. This is a bit surprising.

The only occasion at which q=∆H is when the moles of the limiting reagent in the thermodynamic equation is 1, such that q=∆H/n.

If ∆H is understood to be the molar enthalpy of reaction, you would have to multiply by the amount that reacts, not divide, i.e.

$$q = \Delta_r H \cdot n$$

where $n$ is the amount that reacts (i.e. the change in the amount of any species divided by its stoichiometric coefficient).

What does constant pressure has to do with this equality? Because if we assume such equality of Qp=∆U+P∆V=∆H at constant pressure, then it is also valid at constant volume when ∆V=0 such that Q=∆U=∆H.

That is a different question, and there are good pointers in the comments to learn more. The generally valid equality is:

$$ \Delta H = \Delta U + \Delta (P V)$$

so $P V$ changes when the pressure is constant and the volume changes, or when the volume is constant and the pressure changes (or when both change). This equation refers to enthalpy, not molar enthalpy, as you can figure out from dimensional analysis ($P V$ has dimensions of an energy, not a molar energy).

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  • $\begingroup$ Dear Karsten Theis, I would like to offer you my sincere thanks from the bottom of my heart, I think I understood it know and my confusion has been addressed. Your way of clarifying things in a yet not so complicating manner is peculiar, and I ask god to bless you and your knowledge, because people like you is what we need for constructive knowledge gaining in science. Thank you for using your experience with higher level textbooks to clarify the vagueness of the beginner textbooks when indeed clarity is key to understanding. Once again, thank you so much. $\endgroup$
    – Doe Pual
    Commented Jan 16, 2022 at 9:08

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