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check my answer for a quick standard enthalpy of combustion question

Need your help with this :smile:

Question is calc the standard enthalpy of combustion for ethanol using the following standard enthalpies of formation:

CO2 = -393
H2O = -286
C2H5OH = -278

so balanced equation is CH3CH2OH + 3O2 -> 2CO2 + 3H2O

This is the part im not sure of , is the equation the same as for the formation?

i.e. = Enthalpy of products - Enthalpy of reactants (is there a better way to write this?)

If so, is it [(2 x -393) + (3 x -296)] - [-278] giving -1366

Thanks for the help :smile:
Reply 1
That would be the correct answer, yes. Note though that you're calculating the enthalpy of COMBUSTION by using the data of the enthalpy of FORMATION. So, the enthalpy of combustion is actually the change in the enthalpy of formation.
(edited 10 years ago)
Reply 2
Original post by Doppel
That would be the correct answer, yes. Note though that you're calculating the enthalpy of COMBUSTION by using the data of the enthalpy of FORMATION. So, the enthalpy of combustion is actually the change in the enthalpy of formation.


Thanks for the reply. Could you elaborate on this? Are you saying these two values are always equal? Sorry, im a simpleton :smile:
Reply 3
Yes (I can't think of any exceptions at the moment at least). The enthalpy of combustion is always equal to the change in the enthalpy of formation. You could think about it like this:

The enthalpy of formation is defined as the change in enthalpy in the formation of a compound from its' elements. So, when ethanol is formed, it would've retained heat. When you combust ethanol like in your case, you get CO2 and H2O. Hence, the energy released when you combust ethanol would be the energy needed to form CO2 and H2O subtracted by the energy needed to form ethanol. Obviously, everything has to be balanced.
(edited 10 years ago)
Reply 4
great, thanks for the help :smile:

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