How to find enthalpy of rxn using enthalpies of formation?

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Katie Clark 3B
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Joined: Fri Sep 25, 2015 3:00 am

How to find enthalpy of rxn using enthalpies of formation?

Postby Katie Clark 3B » Sat Jan 23, 2016 5:49 pm

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?

Shreya Banerjee 2F
Posts: 24
Joined: Fri Sep 25, 2015 3:00 am

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

Postby Shreya Banerjee 2F » Sat Jan 23, 2016 6:48 pm

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.

Rachel Lipman
Posts: 45
Joined: Fri Sep 25, 2015 3:00 am

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

Postby Rachel Lipman » Sat Jan 23, 2016 9:39 pm

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."


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