[DKJones96]

I don't remember how to calculate heat of reaction of two compunds. I need the combustion of a mole of Ethanol in this reaction

C2H6O + 3O2 = 3H2O + 2CO2
i know this one gives off 1409.4 kJ of heat but what about the other using O3 in the reaction instead of O2...

C2H6O + 2O3 = 3H2O + 2CO2

any help is appreciated.

[GSR]

hi,

Standard enthalpies of formation are: C2H5OH -228, CO2 -394, and H2O -286 kJ/mol. if you Calculate the enthalpy of the reaction C2H5OH + 3 O2 ----> 2 CO2 + 3 H2O its is (*H = -1418 kJ/mol).

The enthalpy of formation of ozone is 142.7 kJ / mol. The bond energy of O2 is 498 kJ / mol. now we have to think about  the average O=O bond energy of the ozone molecule O=O=O?

The overall reaction is
3 O2 ---> 2 O3   (*H = 286 kJ)
I think now its easy to calculate as per your equation.
** (Note that 3 O=O bonds of oxygen are broken, and 4 O-O bonds of ozone are formed. If the bond energy of ozone is E, then E = (3*498 + 286) kJ / 4 mol   = 445 kJ / mol) .

*** you need to think on how is ozone produced ?

When an oxygen molecule receive a photon (h v), it dissociates into monoatomic (reactive) atoms. These atoms attack an oxygen molecule to form ozone, O3.
O2 + h v ---> O + O
O2 + O ---> O3
The last reaction requires a third molecule to take away the energy associated with the free radical O and O2, and the reaction can be represented by
O2 + O + M ---> O3 + M*
The over all reaction between oxygen and ozone formation is:
3 O2 <--> 2 O3

The absorption of UV light leads to the destruction of ozone
O3 + h v --> O + O2
O3 + O --> 2 O2
A dynamic equilibrium is established in these reactions. The ozone concentration varies due to the amount of radiation received.


I hope this information is  useful.

cheers
GSR