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I am reading a book (Thermodynamics an Engineering approach), I am reading over the section of steady-flow systems with chemical reactions. The book just introduced the following formula for the enthalpy (per mole) of a component as:

$$\bar{H} = \bar{h_f^o} + (\bar{h} - \bar{h^o})$$

Where $\bar{h_f^o}$ is the standard enthalpy of formation at the reference state of 25 ºC and 1 atm, and $\bar{h^o}$ is the sensible enthalpy at the standard reference state of 25 ºC and 1 atm.

How are these two quantities different? In my book they give an example where they calculate the enthalpy for oxygen at 7 ºC, they express it as:

$$\ce{1 kmol O_2 \cdot (0 +\bar{h}_{280K} - \bar{h}_{298K} ) }$$ where $\bar{h}_{280K} $ = 8150 kJ/kmol

and $\bar{h}_{298K} $ = 8669 kJ/mol

I do not quite get how there is a non-zero enthalpy value for oxygen at standard temperature.

Here is a screenshot of the book page that introduces this concept: enter image description here

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    $\begingroup$ The standard enthalpy =0 is for One atmosphere Fugacity. This is equal to the enthalpy of the actual gas extrapolated to zero pressure. The actual enthalpies include deviations from ideality and include intermolecular attractions. Chemists believe in the Ideal Gas Law; engineers know it is a figment $\endgroup$
    – jimchmst
    Commented Jul 30 at 21:59
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    $\begingroup$ Maybe the equation has a typo. $\endgroup$
    – Chet Miller
    Commented Jul 31 at 10:19
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    $\begingroup$ @ChetMiller I could provide a screenshot of the page if that is helpful, they use this form of enthalpy to do calculations in the examples too, really throws me off. $\endgroup$
    – STOI
    Commented Jul 31 at 15:56
  • $\begingroup$ Just added the image guys, best regards. $\endgroup$
    – STOI
    Commented Aug 1 at 22:14

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