## 8.85 part C

sofiakavanaugh
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### 8.85 part C

N2(g) + O2(g)---> 2NO (g) deltaH°=180.6 KJ

c) bomb calorimeter; 492 J absorbed. Mass of N2 oxidized?

How I did this problem was (492J/1 mol rxn)(1 mol rxn/1 mol N2)(1 mol/28.02g)=17.6g

Obviously this is the wrong answer, but can someone please explain to me why this logic is incorrect?
Also, in the solutions manual it divides the 492 KJ absorbed by the deltaH value given to get moles? Im confused as to why this works, is this the moles of N2 in the rxn? Why wouldn't it be the total moles of the rxn?

Thanks :)

Chem_Mod
Posts: 18401
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

### Re: 8.85 part C

Hello, $\Delta H^{\circ}$ is the amount of energy required for the reaction to take place for the formation of 2 moles of NO (190.6 KJ). We are told that 492 J is absorbed. Therefore, using dimensional analysis, you can calculate the number of moles of product and find the mass.

Chem_Mod
Posts: 18401
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

### Re: 8.85 part C

Hello Sofia

Remember that 492 J absorb is the total amount of energy absorbed by oxidizing some unknown amount of nitrogen. In your equation, you are assuming that 492 J comes from reacting 1 mol of N2.
So the way to approach this problem is to find out if reacting one mol of reactant gives you 180.6 kJ (because it is the rxn enthalpy), how much mol of react did you oxidize if you end up getting 492 J released? You can then convert the mol into weight.

Hope it helps!

Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

### Who is online

Users browsing this forum: No registered users and 0 guests