# What is the enthalpy of hydrogenation of propene?

Jan 16, 2017

Normally, we might locate the enthalpy of formation for each substance, but since that for ${\text{C"_3"H}}_{6}$ is not in my textbook, we can use enthalpies of bond formation and bond breaking instead.

Draw out the Lewis structures:

Then you should notice that the reactants overall contain these bonds that are broken:

• $6 \times \text{C"-"H}$ single bonds (${\text{CH}}_{3}$, $\text{CH}$, ${\text{CH}}_{2}$)
• $1 \times \text{C"-"C}$ single bonds
• $1 \times \text{C"="C}$ double bond
• $1 \times \text{H"-"H}$ single bond

and these bonds are made in propane:

• $8 \times \text{C"-"H}$ single bonds (${\text{CH}}_{3}$, ${\text{CH}}_{2}$, ${\text{CH}}_{2}$)
• $2 \times \text{C"-"C}$ single bonds

The relevant bond enthalpies are therefore:

• DeltaH_("C"-"H") = "413 kJ/mol"
• DeltaH_("C"-"C") = "348 kJ/mol"
• DeltaH_("C"="C") = "614 kJ/mol"
• DeltaH_("H"-"H") = "436 kJ/mol"

The enthalpy of reaction can be calculated by keeping track of which bonds were broken and which were made in the reaction. So, from the formula:

color(blue)(DeltaH_"rxn") ~~ sum_R n_RDeltaH_"break" - sum_P n_PDeltaH_"form"

= [6DeltaH_("C"-"H") + 1DeltaH_("C"-"C") + 1DeltaH_("C"="C") + 1DeltaH_("H"-"H")] "kJ" - [8DeltaH_("C"-"H") + 2DeltaH_("C"-"C")] "kJ"

$= \left[\left(\text{6 mol")("413 kJ/mol") + ("1 mol")("348 kJ/mol") + ("1 mol")("614 kJ/mol") + ("1 mol")("436 kJ/mol")] - [("8 mol")("413 kJ/mol") + ("2 mol")("348 kJ/mol}\right)\right]$

$= \textcolor{b l u e}{- \text{124 kJ/mol}}$