[mahela007]

Hi there... this is my first post here.
I'm trying to understand the concept of enthalpy of formation but I've got stuck.
I have understood that since it is impossible to determine the actual enthalpy content of a substance, we need to define an arbitrary reference point as 0. Therefore, the standard enthalpy content of any element (in it's standard state)  is designated as 0. This seems to say that all elements have equal internal enthalpy in their standard states (such as graphite and O2) How can this be? How can O2 have the same internal enthalpy as graphite? If they don't have the same internal enthalpy, then why do we call the internal enthalpy of both 0? A good answer will be well appreciated. ;-)

(my text book says that the normal definition of standard enthalpy of formation is derived from the fact that the enthalpy of all elements is designated as 0... not the other way round. That is to say, thinking of O2 and graphite as having an enthalpy of 0 is what allows us to form that definition. )

[Borek]

we need to define an arbitrary reference point as 0

If they don't have the same internal enthalpy, then why do we call the internal enthalpy of both 0?

Because that's our arbitrary decision?

Enthalpy is a function of state, that (between other things) means reference point doesn't matter, as we are always dealing with changes between initial and final state. It is not that internal enthalpy of both IS 0 - it doesn't matter what the real value is. It is just like stepping up a 2 feet chair - doesn't matter if the chair stands on the first floor or on the tenth floor - difference between the initial and final state is always 2 feet.

[Zeppos10]

I have understood that since it is impossible to determine the actual enthalpy content of a substance, we need to define an arbitrary reference point as 0. 

This is a common belief [urban myth?] in chemistry, but in 1996 Spencer,  in "An Approach to reaction thermodynamics through enthalpies, entropies and free energies of atomization." (JCE 73:631) has shown otherwise. The absolute enthalpy of O2 is -492kJ/mol: this can be obtained by assigning a value of 6.2 to the enthalpy of atomic oxygen gas (O) at 298K and 1 atm with Habs=5RT/2, and using the enthalpy of atomization to get the enthalpy of O2. 

[Borek]

this can be obtained by assigning a value of 6.2 to the enthalpy of atomic oxygen gas (O) at 298K and 1 atm

What if I assign 3.19?

[Zeppos10]

What if I assign 3.19?

That would be fine with me if this choice (3.19) is well-motivated.
However an ideal gas (of O) at 298K and 1 atm has an internal energy U=3.7 kJ/mol (=3RT/2) and a pV-energy of 2.5 kJ/mol (=pV=1RT), together 6.2 kJ/mol.

[Borek]

O is not an ideal gas. 6.2 is just an approximation.

Besides, as far as I can tell, word "absolute" is not even used in the paper you cited. Are you sure you are not trying to fight a "myth" with another myth?

[Zeppos10]

O is not an ideal gas. 6.2 is just an approximation.

Yes, 6.2 is an approximation and stands for all values between 6.15 and 6.25.
The exact value for the enthalpy of all (mono-atomic) ideal gases, taking 5 significant numbers for R and T, is: H(abs)=6.1974 kJ/mol.
For non-ideal gases this would be an upper limit. My estimate for the true value would be 6.17 ~ 6.18 kJ/mole.
But, rather then worrying about the true absolute value it would be more usefull to assign a value of 6.20 to all mono-atomic gases, and be satisfied that the resulting values for H are approximately absolute, or absolute within the limits of accuracy that apply the H values for chemical reactions. 

[Borek]

Obviously you have selected to ignore fact that the paper you quote doesn't address the problem?

Any reason we shoudl trust you if you started with false information?

[Zeppos10]

Just read the above with the proper amount of skepsis: The article by Spencer has shown to ME how enthalpy can be approached differently and how to arrive at an enthalpy value that is absolute for all practical purposes.
The original question was: How can O2 have the same (internal?) enthalpy as graphite? I hope that the readers now understand that they do not, unless they are assigned equal values quite arbitrarily.

[Borek]

OK, I got the point. I still don't feel convinced that your approach is better, but that's another story.

[Zeppos10]

OK, I got the point. I still don't feel convinced that your approach is better, but that's another story.

Here is a reference of somebody who feels convinced:
KB Kester writes in a book review (The Chemical Educator, 3(1998)):
"I prefer enthalpies of atom combination because they are directly related to bond forming and bond breaking, and because having the zero state in enthalpies of atom combination be the free gaseous atom of each element is both simpler and more coherent // With enthalpies of atom combination, the zero state is the same for all(!) elements: the element in its monoatomic gaseous state."