If given 2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g) and DeltaH = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?

1 Answer
Sep 9, 2016

DeltaH""^@""_"combustion" = -2875*kJ*mol^-1

Explanation:

The "molar enthalpy of combustion" is the energy associated with the complete combustion of 1 mole of hydrocarbon to give carbon dioxide and water. The given equation quotes enthalpy output,

"PER MOLE OF REACTION AS WRITTEN"

And thus it represents TWICE the enthalpy of combustion of butane because TWO moles of butane were combusted.

DeltaH""^@""_"combustion" = (-5750*kJ*mol^-1)/2= -2875*kJ*mol^-1

Of course, certain reference standards are specified (the most annoying of which is 100 kPa, but that's another story!).