Question

What is the enthalpy of hydrogenation?

Answer 1

Well, it is just like any other reaction... you could simply find the enthalpy of reaction for a particular hydrogenation reaction (i.e. addition of `"H"_2(g)` across a `pi` bond), such as...

The standard enthalpy of hydrogenation of ethene would be gotten from...

`DeltaH_(rxn)^@ = sum_P nu_P DeltaH_(f,P)^@ - sum_R nu_R DeltaH_(f,R)^@`

where:

• `nu` is the stoichiometric coefficient.
• `R` and `P` stand for reactant and product, respectively.
• `DeltaH^@` is the standard molar enthalpy.

With enthalpy of formation data obtained from NIST, we have:

`DeltaH_(f,"C"_2"H"_4(g))^@ = "52.47 kJ/mol"`

`DeltaH_(f,"H"_2(g))^@ = "0.00 kJ/mol"` (why?)

`DeltaH_(f,"C"_2"H"_6(g))^@ = -"83.8 kJ/mol"`

We get...

`color(blue)(DeltaH_(hydr)^@)`

`= [nu_(C_2H_6(g))DeltaH_(f,"C"_2"H"_6(g))^@] - [nu_(C_2H_4(g))DeltaH_(f,"C"_2"H"_4(g))^@ + nu_(H_2(g))DeltaH_(f,"H"_2(g))^@]`

`= ["1 equiv." xx overbrace(-"83.8 kJ/mol")^(DeltaH_(f,"C"_2"H"_6(g))^@)] - ["1 equiv." xx overbrace("52.47 kJ/mol")^(DeltaH_(f,"C"_2"H"_4(g))^@) + "1 equiv." xx overbrace("0.00 kJ/mol")^(DeltaH_(f,"H"_2(g))^@)]`

`= color(blue)(-"136.27 kJ/mol")`