How to calculate the molar formation of propanol?
If the molar enthalpy of combustion of propanol is -1843.7 kj/mol, what is the molar enthalpy of formation of propanol?
2C3H7OH + 9O2 -> 6CO2 + 8H2O
I used: (sum of products) - (sum of reactants) = enthalpy of formation. However, i don't understand why I used the molar enthalpies of FORMATION of water and CO2, but then I had to use the molar enthalpy of COMBUSTION for propanol, how can I just mix molar enthalpies in an equation like that. But somehow that's how I got the answer.
Answers & Comments
Go back and learn your definitions of enthalpies!
In your equation, the enthalpy change is for combustion of propanol. It is NOT same as the formation of propanol.
As the definition says...
"Enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states and under standard conditions of temperature and pressure."
Now, if you set up Hess's cycle as shown below
2C3H7OH + 9O2 ---delta H combustion of propanol---> 6CO2 + 8H2O (final products)
delta H formation of propanol
6C + 4H2 + 10O2 (initial reactants)
and label an arrow going from initial reactants to propanol as delta H formation of propanol and another arrow going from initial reactants to final products as delta H formation of water and carbon dioxide (this arrow not shown/labelled) then applying Hess's law
delta H formation of propanol + delta H combustion of propanol = delta H formation of water and carbon dioxide.
Rearrange this equation for delta H formation of propanol, and you will see why!