2
$\begingroup$

My question is somewhat related to this question. When I look at a table of standard values there are two entries for Hydrogen:

  • $\ce {H2(g)} $ - 0 kJ/mol
  • $\ce {H(g)} $ - 218 kJ/mol

I understand the $\ce{H2(g)} $ is 0 because it is used as a reference, why is monatomic $\ce{ H(g)}$ given in the table and what does it mean? When would one use it in a calculation of enthalpy change?

From what I understand Hydrogen only exists as a monatomic gas at very high temperatures, the Standard enthalpies of formation are given at 1 bar, 298.15 K.

Improve this question
$\endgroup$
1
  • $\begingroup$ The value for the hydrogen atom is one-half the bond dissociation energy of the hydrogen molecule. $\endgroup$
    – user55119
    Commented Dec 9, 2021 at 0:56

1 Answer 1

Reset to default
2
$\begingroup$

The value may be used whenever atomic hydrogen is present as a (probably intermediate) species. E.g. take the reaction

$$\ce{2H <=> H2}$$

which is the basic reaction in atomic hydrogen welding.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ This is correct - often the relevant species is a lone proton. For example, in the formation of solid metal hydrides you need H to enter, diffuse, and react with the metal. H$_{2}$ just won't do it. $\endgroup$
    – Jon Custer
    Commented Apr 18, 2016 at 17:20
  • 1
    $\begingroup$ @JonCuster does that mean that we use the $\ce{H(g)}$ in Hess's law when we are looking at, for instance, the combustion of hydrogen with oxygen, since at some point monatomic hydrogen is an intermediate specie in that reaction? $\endgroup$
    – samuelschaefer
    Commented Apr 20, 2016 at 5:08
  • 1
    $\begingroup$ Well, you get to choose your standard state, but you have to be consistent. The end result is the same, how you get there is slightly different. $\endgroup$
    – Jon Custer
    Commented Apr 20, 2016 at 12:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.