What is the enthalpy of hydrogenation of propene?

Normally, we might locate the enthalpy of formation for each substance, but since that for #"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:

• #6xx"C"-"H"# single bonds (#"CH"_3#, #"CH"#, #"CH"_2#)
• #1xx"C"-"C"# single bonds
• #1xx"C"="C"# double bond
• #1xx"H"-"H"# single bond

and these bonds are made in propane:

• #8xx"C"-"H"# single bonds (#"CH"_3#, #"CH"_2#, #"CH"_2#)
• #2xx"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"#

#= [("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")]#

#= color(blue)(-"124 kJ/mol")#