7.71  [ENDORSED]

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Alwel7041
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7.71

Postby Alwel7041 » Fri Jan 10, 2014 10:03 pm

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

Postby Chem_Mod » Fri Jan 10, 2014 10:58 pm

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
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Re: 7.71

Postby Robert » Sun Jan 12, 2014 12:02 pm

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
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Re: 7.71

Postby mitalisharma2B » Mon Jan 22, 2018 12:10 pm

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?

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Re: 7.71  [ENDORSED]

Postby Chem_Mod » Mon Jan 22, 2018 2:30 pm

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.

Hadji Yono-Cruz 2L
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Re: 7.71

Postby Hadji Yono-Cruz 2L » Mon Feb 11, 2019 3:04 pm

The question is asking what is the standard enthalpy formation of N2O5, not the overall reaction. From the chemical equation, O2 is in its natural state, so the enthalpy for O2 is 0. We are also given the standard enthalpy of formation for NO which is 90.25 KJ. With this information we know that delta H of the reaction= -114.1 and delta H= sum of the enthalpies of the products minus the sum of the enthalpies of the reactants. We need to solve for NO, so set the equation as -114.1= 2n - 2(90.25). The n will give you the standard enthalpy formation of NO2 which is around 33.2 KJ. You can use this same method for the next equation because you know the standard enthalpy formation of NO2, O2, and the overall reaction. Now solve for N2O5 with the equation: -110.2= 2n - 4(33.2). N should be around 11.3 KJ and that is the standard enthalpy formation of N2O5.


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