Question

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?

Answer 1

`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!).