## 7.71 [ENDORSED]

Alwel7041
Posts: 2
Joined: Fri Sep 20, 2013 3:00 am

### 7.71

Calculate the Standard Enthalpy of Formation of Dinitrogen pentoxide from the following data and from the standard enthalpy of formation of nitric oxide:
2NO(g) + O2(g) --> 2NO2(g) H=-114.1kJ
4NO2(g) + O2(g) -->2N2O5(g) H= -110.2 kJ

For this question, I do not understand how to figure out the "overall" reaction: the formation of dinitrogen pentoxide.

Chem_Mod
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### Re: 7.71

Since N2 and O2 have been assigned an arbitrary heat of formation of 0. These can both be used as starting materials for a "reaction" forming dinitrogen pentoxide.

Robert
Posts: 144
Joined: Fri Sep 20, 2013 3:00 am

### Re: 7.71

If you mean delta Hrxn (overall heat of the reaction), you would multiply the coefficients of the first equation by 2. By doing so, you must also multiply the delta H of that reaction. When you add the two equations together, the NO2 cancels out on both sides. Also add the heats of the reactions together; the overall heat is something like -338.4 kJ.

Overall equation 4NO + 3O2 ---> 2N2O5 delta Hrxn= -338.4 kJ

Then create the equation where the sum of the [heat of formation of the products] minus the sum of the [heat of formation of the reactants] equals -338.4kJ. The heat of the formation of the product is basically dinitrogen pentaoxide. so solve for the variable in the equation. You can find the heat of formation of the reactant NO in appendix 2A.

mitalisharma2B
Posts: 39
Joined: Fri Sep 29, 2017 7:07 am

### Re: 7.71

Why is it that we must create an "equation where the sum of the [heat of formation of the products] minus the sum of the [heat of formation of the reactants] equals -338.4kJ"?

Why wouldn't the answer be -338.4kJ?

Chem_Mod
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### Re: 7.71  [ENDORSED]

That is not the answer because the starting materials are not in their elemental states. Because of this, you would need to subtract the enthalpies of formations of the reactants from the values to find the enthalpy of formation of just the product.