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

1 Answer
Truong-Son N.
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.

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