Question

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)`

Answer 1

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.

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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.