What is the enthalpy of hydrogenation of propene?

1 Answer
Jan 16, 2017

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")