8.65 and 8.67

Moderators: Chem_Mod, Chem_Admin

Emily Glaser 1F
Posts: 156
Joined: Thu Jul 27, 2017 3:01 am

8.65 and 8.67

Postby Emily Glaser 1F » Tue Jan 16, 2018 11:56 pm

In both of these homework problems, the solution manual provides necessary equations out of thin air:

for 8.65, N2(g) + (5/2)O2(g) --> N2O5(g)

for 8.67, H2(g) + (1/2)O2(g) --> H2O(l)

Where do these equations come from? For question 8.65, I found the correct answer without using that equation. I wasn't even sure how to start 8.67 without the solution manual. Help

Michelle Steinberg2J
Posts: 73
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.65 and 8.67

Postby Michelle Steinberg2J » Wed Jan 17, 2018 12:16 am

These are just the overall equations that get you to the necessary product. That is, you must alter the equations provided in 65 to get to that overall equation. To do that, you can multiply the second equation by 1/2. This will allow 2 NO2 to cancel, and we want this to cancel because it is not included in the equation to reach the product.

Hope this helps!

Justin Chu 1G
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

Re: 8.65 and 8.67

Postby Justin Chu 1G » Wed Jan 17, 2018 10:39 am

Conversely you could also multiply the first equation by 2 to get 4 NO + 2 O2 --> 4 NO2. In this case you would also have to multiply the deltaH by 2. Then the 4 NO2s would cancel out leaving you with the first equation being 4 NO + 2 O2 --> and the second equation with O2 --> 2 N2O5. You would then add the two equations together to get 4 NO + 3 O2 --> 2 N2O5 with a deltaH of -338.4 kJ. You would then find deltaH for one mol of N2O5 rather than 2.

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

Who is online

Users browsing this forum: No registered users and 1 guest