# What is the enthalpy of hydrogenation?

##### 1 Answer

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

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")#