You usually calculate the change of combustion from enthalpies of formation.
The standard enthalpy of combustion is ##ΔH_”c”^°##.
It is the heat evolved when 1 mol of a substance burns completely in oxygen at standard conditions. For example,
##”C”_2″H”_2″(g)” + 5/2″O”_2″(g)” → “2CO”_2″(g)” + “H”_2″O(l)”##
You calculate ##ΔH_”c”^°## from standard enthalpies of formation:
##ΔH_”c”^o = ∑ΔH_”f”^°”(p)” – ∑ΔH_”f”^°”(r)”##
where ##”p”## stands for “products” and ##”r”## stands for “reactants”.
For each product, you multiply its ##ΔH_”f”^°## by its coefficient in the balanced equation and add them together.
Do the same for the reactants. Subtract the reactant sum from the product sum.
EXAMPLE:
Use the following enthalpies of formation to calculate the standard enthalpy of combustion of acetylene, ##”C”_2″H”_2##.
##”C”_2″H”_2″(g)” + 5/2″O”_2″(g)” → “2CO”_2″(g)” + “H”_2″O(l)”##
##DeltaH_(“C”_2″H”_2″(g)”)^o = “226.73 kJ/mol”##; ##DeltaH_(“CO”_2″(g)”)^o = “-393.5 kJ/mol”##;
##DeltaH_(“H”_2″O(l)”)^o = “-285.8 kJ/mol”##
Solution:
##”C”_2″H”_2″(g)” + 5/2″O”_2″(g)” → “2CO”_2″(g)” + “H”_2″O(l)”##
##ΔH_”c”^o = ∑ΔH_”f”^°”(p)” – ∑ΔH_”f”^°”(r)”##
##”[2 × (-393.5) + (-295.8)] – [226.7 + 0] kJ” = “-1082.8 – 226.7” =##
##”-1309.5 kJ”##
The heat of combustion of acetylene is -1309.5 kJ/mol.
Here is a video that discusses how to calculate the enthalpy change when 0.13 g of butane is burned.
Video from: Noel Pauller
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