Chapter 8 #65

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Michael Lonsway 3O
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Joined: Wed Sep 21, 2016 2:57 pm

Chapter 8 #65

Postby Michael Lonsway 3O » Sun Jan 15, 2017 1:15 pm

Calculate the standard enthalpy of formation of dinitrogen pentoxide from the following data,
and from the standard enthalpy of formation of nitric oxide, NO (see Appendix 2A)
2NO(g) + O2(g) ---> 2NO2(g) Delta h= -114.1kJ
4NO2(g) + O2(g) ---> 2N2O5(g) Delta h = -110.2kJ

I'm having trouble getting the delta h of dinitrogen pentoxide using the additive method of enthalpies for this problem

Diwana Lucero 3K
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Joined: Fri Jul 15, 2016 3:00 am

Re: Chapter 8 #65

Postby Diwana Lucero 3K » Sun Jan 15, 2017 2:02 pm

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For this problem, you want to add the first equation to half of the second equation. But since we only want the standard enthalpy of formation of N2O5, you use the standard enthalpy of formation equation to find the value for the standard enthalpy of N2O5.

Michael Lonsway 3O
Posts: 43
Joined: Wed Sep 21, 2016 2:57 pm

Re: Chapter 8 #65

Postby Michael Lonsway 3O » Sun Jan 15, 2017 2:34 pm

I'm confused about the second part of the solution. I understand that we needed to find the enthalpy of formation of N2O5 but why is -169.2kJ not the final answer?

Aleena_Sorf_2A
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Joined: Sat Jul 09, 2016 3:00 am

Re: Chapter 8 #65

Postby Aleena_Sorf_2A » Sun Jan 15, 2017 2:50 pm

The value -169,2 kJ is not the answer because the final equation does to contain NO as a a reactant. In order to cancel the unwanted NO, we must do the following:
-169.2kJ= the enthalpy of formation of N2O5 (which is what we are trying to solve for) - 2(+90.25 kJ) = +11.3kJ
The second part of this equation above is the number of moles of NO x the enthalpy of formation of NO.

Madeline_Foo_3J
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Joined: Wed Sep 21, 2016 3:00 pm

Re: Chapter 8 #65

Postby Madeline_Foo_3J » Sun Jan 15, 2017 3:57 pm

How would you have known to half the second equation ?

andrea_Disc3D
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Joined: Wed Sep 21, 2016 2:55 pm

Re: Chapter 8 #65

Postby andrea_Disc3D » Sun Jan 15, 2017 4:52 pm

Madeline_Foo_3J wrote:How would you have known to half the second equation ?


You know you have to half the second equation because in order to form dinitrogen pentoxide (N2O5) you need 2NO which the first equation gives you. But the first equation also produces 2NO2 which you do not want in your final reaction equation.
Therefore, in order to cancel out this product (2NO2) and just get N2O5 for your product you need to have 2NO2 as a reactant. The second equation gives you 4NO2 as an reactant, but you only want half of this amount (or else you will have 2NO2 (as a reactant) included in your final reaction equation. So, you multiply the second equation by half to cancel the product of the first equation and to get 1 mol of N2O5 (instead of 2 mol) which is what you want anyways.

Overall, you just need to keep in mind the final reaction you want and manipulate the other equations to get your desired reaction.

I hope this makes sense!


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