Which energy change corresponds to the enthalpy change of atomization of H_2 at 298 K ?

1 Answer
Apr 15, 2018

Atomization is simply the breaking of bonds to yield "1 mol" of the atoms in the gas phase. So, if you have

"H"_2(g) -> 2"H"(g),

then you have twice DeltaH_"atom"^@, i.e. you've broken one "mol" of "H"-"H" bonds for every one "mol" of "H"_2(g) molecules to form two "mols" of "H" atoms.

DeltaH_"atom"^@ is based on "1 mol" of atoms, so really represents:

1/2"H"_2(g) -> "H"(g)

So, DeltaH_"atom"^@ is half the bond energy of a "H"-"H" bond.