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How to find enthalpy of rxn using enthalpies of formation?

Posted: Sat Jan 23, 2016 5:49 pm
by Katie Clark 3B
On the homework problem 8.51, the reaction is 4C7H5N3O6+21O2 -> 28CO2+ 10H2O+6N2. It is also given that the enthalpy of formation for TNT is -67 kJ per mol. Can someone explain why the solutions manual only used the enthalpies of formation for TNT, H2O and CO2 to find the enthalpy of reaction. Shouldn't it include all the products and all the reactants?

Re: How to find enthalpy of rxn using enthalpies of formatio

Posted: Sat Jan 23, 2016 6:48 pm
by Shreya Banerjee 2F
It does include all the reactants and products, but the rest of them are elements. The molar enthalpy of an element is 0, so it doesn't have to be included in the calculation. The book solution omitted nitrogen and oxygen because their enthalpy is 0 and wouldn't contribute anything to the answer.

Re: How to find enthalpy of rxn using enthalpies of formatio

Posted: Sat Jan 23, 2016 9:39 pm
by Rachel Lipman
I was just reviewing this question as well, so basically because O2 and N2 are not included in the overall calculation of enthalpy because "by definition, the standard enthalpy of formation of an element in its most stable form is zero."