# 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?

Sep 9, 2016

$\Delta H \text{^@""_"combustion}$ $=$ $- 2875 \cdot k J \cdot m o {l}^{-} 1$

#### Explanation:

The $\text{molar enthalpy of combustion}$ 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,

$\text{PER MOLE OF REACTION AS WRITTEN}$

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

$\Delta H \text{^@""_"combustion}$ $=$ $\frac{- 5750 \cdot k J \cdot m o {l}^{-} 1}{2} = - 2875 \cdot k J \cdot m o {l}^{-} 1$

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