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The standard enthalpy of formation $\Delta H_f^°$ of pure elements is zero by definition. Why is that a convention? It is true that enthalpy is defined unless a constant (like energy and entropy), but enthalpy of formation is actually a variation of enthalpy, so we don't really care.

Also, how could we possibly choose something else than zero if there's no heat transfer?

Note: I'm using General Chemistry (Ralph Petrucci)

Mr. Feynman
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  • Entropy is absolute. No standards required. Energy, as you note, does require a reference. I am hard pressed to think of a reference besides elements in their standard state that would be so simple, universal, and reasonably self-consistent. – Zhe Aug 25 '20 at 14:23
  • Energy definitely needs a reference but energy change does not. – Mr. Feynman Aug 25 '20 at 14:26
  • Well, the enthalpy of formation of pure elements from pure elements is indeed zero. – Poutnik Aug 25 '20 at 15:07
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  • Enthalpy is similar to gravific energy. It needs a reference. For gravific energy, the zero is arbitrary, and it is at the sea level. For enthalpy considerations, it is also arbitrary, The zero has been once decided to be defined by the pure elements. Other choices could have been possible, like the situation where all atoms are separated from one another. But the scientists all agreed to accept that the zero enthalpy is obtained when the elements are in their pure state at ordinary temperature and pressure. – Maurice Aug 25 '20 at 15:56
  • Zero enthalpy is not the same as zero enthalpy of formation, as isobarically, dH=n.Cp.dT. – Poutnik Aug 25 '20 at 16:34
  • Maybe the problem here is how I'm reading your question. The enthalpy of formation of elements is by definition zero because the reaction is trivial (e.g., $\ce{H2 -> H2}$) and since the reactant is, by definition, pegged to zero, the delta is also zero. – Zhe Aug 25 '20 at 18:58
  • The problem is that enthalpy is defined unless an arbitrary constant, but that constant elides when working with entropy change, so I'm saying you can't just define some arbitrary value for it. It is naturally zero because no energy is exchanged (like in the trivial reaction you wrote), we are not setting it to be zero (this is what it looks like to me). – Mr. Feynman Aug 25 '20 at 19:04
  • The definition is something like 'formation of pure elements in their normal physical state at 1 atm and at 298.15 K' (and may now be 1 bar or a different temperature). But the important point is normal physical state so some are solids, some liquids and some gasses. So making all these zero saves a lot of extra fiddling about compared to that if we chose, say, free atoms at some huge temperature as the reference. For example consider then the calculation forming solid carbon or gaseous Cl2. – porphyrin Aug 26 '20 at 11:54

1 Answers1

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It starts with this: The standard enthalpy of formation of any substance, $\Delta H^\circ_\mathrm{f}$, is (by convention) defined as the enthalpy change for the reaction at 1 bar and a specified temperature (usually $\pu{298.15 K}$), in which the product is 1 mole of that substance, and the reactants are its component elements in their respective standard states.

For instance, $\Delta H^\circ_\mathrm{f}$ for $\ce{H2O_{(l)}}$ is equal to $\Delta H^\circ$ for the following reaction at standard state:

$$\ce{H2_{(g)} + 1/2O2_{(g)}->H2O_{(l)}},$$

since $\ce{H2_{(g)}}$ and $\ce{O2_{(g)}}$ are the respective standard states for hydrogen and oxygen.

Once you've accepted this definition as the convention for determining $\Delta H^\circ_\mathrm{f}$ for any substance, it directly follows that the value of $\Delta H^\circ_\mathrm{f}$ for any element in its standard state must be zero. For instance, $\Delta H^\circ_\mathrm{f}$ for $\ce{H2_{(g)}}$ is equal to $\Delta H^\circ$ for the following reaction:

$$\ce{H2_{(g)} -> H2_{(g)}},$$

which is necessarily zero.

To give an analogy: Suppose you define the "altitude of formation", $\Delta z_f$, of any location on earth as the change in altitude necessary to reach that location from sea level.

Hence $\Delta z_f$ for the summit of Mt. Everest is $\Delta z$ for the altitutude change associated with:

$$\ce{sea level -> summit of Everest},$$

which is 29,029 feet.

It necessarily follows, from this convention, that $\Delta z_f$ for any location at sea level is zero, since that would be equal to $\Delta z$ for:

$$\ce{sea level -> sea level}$$.

theorist
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    Ooooh! Thank you so much, this is what I've been looking for so far. Maybe I was misinterpreting the other answers, though. I'll accept your answer because I consider it to be the clearest one. The key word was "once you have accepted this convention, it directly follows..." that is coherent with how I expected it to be. Some books, after defining the enthalpy of formation, say we choose (not that it directly follows) the zero for pure elements as a convention. I suppose that was bad phrasing or maybe mistranslations in the italian edition. – Mr. Feynman Aug 27 '20 at 09:38