Problem 8.65

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Chin_Alyssa_3I
Posts: 34
Joined: Fri Jul 22, 2016 3:00 am

Problem 8.65

Postby Chin_Alyssa_3I » Fri Jan 13, 2017 6:14 pm

8.65 Calculate the standard enthalpy of formation of dinitrogen pentoxide from the following data,

2 NO(g) + O2 (g) -> 2NO2(g)  H°= -114.1 kJ
4 NO2(g) + O2(g) -> 2 N2O5(g) H°= -110.2 kJ

and from the standard enthalpy of formation of nitric oxide, NO (see Appendix 2A).

At first I approached it using Hess's Method, but the equations don't add up. What approach should I use? What is my first step?

kara_kremer_2N
Posts: 20
Joined: Fri Jul 15, 2016 3:00 am

Re: Problem 8.65

Postby kara_kremer_2N » Fri Jan 13, 2017 7:07 pm

the desired reaction is: N2(g)+5/2 O2(g)--> N2O5(g)
Divide the second reaction by 2 and add the first and second together to get: 2 NO(g)+3/2 O2(g)--> N2O5(g) with an enthalpy of-169.2 kJ.

Then calculate the standard enthalpy of formation

Marisa_Woo_2G
Posts: 24
Joined: Wed Sep 21, 2016 2:56 pm

Re: Problem 8.65

Postby Marisa_Woo_2G » Fri Jan 13, 2017 7:16 pm

Hi!

To my understanding, I solved this problem with Hess's law for the first part and then the method #3 with the standard heats of formation. First, I set up the final chemical equation we are trying to find, the formation of dinitrogen pentaoxide: The final equation is: N2 (g) + 5/2 O2 (g) --> N2O5

You would then add the two chemical equations given in the problem. You would look at which equation has the compounds you want on the product side and which you want on the reactant side. Since the second equation has 2 moles of N2O5, you would divide the equation and the enthalpy by 2 in order to get the one mole of N2O5 you want in the final equation. The first chemical equation has the reactant O2, so the equation would remain the same when adding the two chemical equations.

The equation you would get from adding this would be 2NO (g)+ 3/2 O2 (g) --> N2O5 and you would add the enthalpies of the equations. This would give you -169.2 kJ. The solution manual says that this is the enthalpy of formation of N2O5 (g) minus twice the enthalpy of formation of NO. What this means is that they added the heat of formation, which is from method 3. This is products minus reactants. If we look at the equation from when we added, 2NO (g)+ 3/2 O2 (g) --> N2O5, there are 2 moles of NO (the standard enthalpy of formation of nitric acid/NO is 90.25 kJ), 3/2 moles of O2 is a free element (meaning the standard enthalpy of formation of molecular oxygen is 0 kJ), and one mole of N2O5 (which is what we are solving for).

So we would set up the equation: -169.2 kJ = sum of products-sum of reactants = (the enthalpy of formation of N2O5) - (2(90.25kJ)+0). You would then get 11.3 kJ.

Not too sure if I have all my reasoning right, but hope this helps!

deantuazon2G
Posts: 13
Joined: Fri Jun 17, 2016 11:28 am

Re: Problem 8.65

Postby deantuazon2G » Sat Jan 14, 2017 4:46 pm

2 NO(g) + O2 (g) -> 2NO2(g) H°= -114.1 kJ
4 NO2(g) + O2(g) -> 2 N2O5(g) H°= -110,2

First, I divided the second equation by 2 to get 2NO2(g) + 1/2O2(g) --> N2O5(g) with a dH= -55.2. Then I added the equations together to get 2NO(g) + 3/2O2(g) --> N205(g) with a dH= -169.2. The equation for NO is N2(g) + O2(g)--> 2NO (g) with a dH= 2(90.25)=180.5 kJ. Lastly, add these last two chemical equations together to get N2(g) + 5/2O2(g) --> N2O5(g) which is the product we've been looking for. We then add -169.2+180.5 to get 11.3 kJ.


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