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

Answer:

#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!).