Question

What is enthalpy of hydrogenation?

Answer 1

It's just the heat/enthalpy of reaction for a hydrogenation reaction.

One example is:

`"H"_2"C"="CH"_2 + stackrel("H"_2)(->) " H"_3"C"-"CH"_3`
`color(white)(aaaaaaaaaaaaa)^"Pd/C"`

`color(blue)(DeltaH_"rxn"^@ ~~ -"136 kJ/mol")`

You can determine this from the standard enthalpies of formation.

`\mathbf(DeltaH_"rxn"^@ = sum_P n_PDeltaH_f^@ - sum_R n_RDeltaH_f^@)`

where `R` is reactants, `P` is products, `n` is the stoichiometric coefficient, `H` is enthalpy, and `@` labels the enthalpy as the "standard" enthalpy (`T = 25^@"C"` and `P = "1 atm"`). You should be able to just look `DeltaH_f^@` up in your textbook appendix.

`DeltaH_(f, "H"_2"C"="CH"_2(g))^@ = "52.3 kJ/mol"`

`DeltaH_(f, "H"_2(g))^@ = "0 kJ/mol"`

`DeltaH_(f, "H"_3"C"-"CH"_3(g))^@ = "-83.85 kJ/mol"`

Note that the catalyst in the reaction example does not numerically contribute to the enthalpy of hydrogenation.

So, evaluating this is pretty simple:

`color(blue)(DeltaH_"rxn"^@) = (n_("H"_3"C"-"CH"_3(g))DeltaH_(f, "H"_3"C"-"CH"_3(g))^@) - (n_("H"_2"C"="CH"_2(g))DeltaH_(f, "H"_2"C"="CH"_2(g))^@ + n_("H"_2(g))DeltaH_(f, "H"_2(g))^@)`

`= (1*"-83.85 kJ/mol") - (1*"52.3 kJ/mol" + 1*"0 kJ/mol")`

`= color(blue)(-"136.15 kJ/mol")`