What is the enthalpy of hydrogenation of propene?
1 Answer
Normally, we might locate the enthalpy of formation for each substance, but since that for
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 bonds1xx"C"="C" double bond1xx"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")