# What is the enthalpy of hydrogenation?

##### 1 Answer
Jul 29, 2017

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 ${\text{H}}_{2} \left(g\right)$ across a $\pi$ bond), such as...

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

$\Delta {H}_{r x n}^{\circ} = {\sum}_{P} {\nu}_{P} \Delta {H}_{f , P}^{\circ} - {\sum}_{R} {\nu}_{R} \Delta {H}_{f , R}^{\circ}$

where:

• $\nu$ is the stoichiometric coefficient.
• $R$ and $P$ stand for reactant and product, respectively.
• $\Delta {H}^{\circ}$ 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...

$\textcolor{b l u e}{\Delta {H}_{h y \mathrm{dr}}^{\circ}}$

$= \left[{\nu}_{{C}_{2} {H}_{6} \left(g\right)} \Delta {H}_{f , {\text{C"_2"H"_6(g))^@] - [nu_(C_2H_4(g))DeltaH_(f,"C"_2"H"_4(g))^@ + nu_(H_2(g))DeltaH_(f,"H}}_{2} \left(g\right)}^{\circ}\right]$

= ["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))^@)]

$= \textcolor{b l u e}{- \text{136.27 kJ/mol}}$