The standard enthalpy of formation of #H_2(g)# is zero. But sometimes we do write( which is also given in my book) #2H(g) -> H_2#; #DeltaH = -"435 kJ/mol"#. This surely suggests that the enthalpy of formation of #H_2# is non-zero?

Then why do we take the enthalpy of formation of #H_2#(g) as zero? Shouldn't it be #DeltaH = -(435KJ)/(mol)#

1 Answer
Truong-Son N.
Jan 6, 2018

Because it is still zero by how formation reactions are defined. You have written a reaction that quantifies forming an #"H"-"H"# bond, the reverse of an atomization reaction for a gaseous reactant.

What you have written is not formation of #"H"_2(g)# from its elemental state, which is required by conventional formation reactions. You have written how two #"H"# atoms bond to form #"H"_2(g)#, and #"H"(g)# is not the elemental state of hydrogen gas.

The formation reaction for something nontrivial is water gas, let's say:

#color(blue)("H"_2(g)) + 1/2"O"_2(g) -> "H"_2"O"(g)#

It is not, however,

#color(red)(2"H"(g)) + 1/2"O"_2(g) -> "H"_2"O"(g)#

Similarly, the formation reaction for #"H"_2(g)# is not

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

but

#"H"_2(g) -> "H"_2(g)#.

And so, the enthalpy of formation of #"H"_2(g)# is always zero.

Related questions
Impact of this question
152 views around the world
You can reuse this answer
Creative Commons License