Question 8.55

Moderators: Chem_Mod, Chem_Admin

Laura Riccardelli
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

Question 8.55

Postby Laura Riccardelli » Thu Jan 11, 2018 12:57 pm

In question 8.55 in the answer manual, in order to get ride of the 1.5 O2 in the first equation to end up with the second equation the O2 molecules in the second equation is made into 1.5 O2 as well but nothing is changed in the rest of the equation. I am wondering how you are able to change the O2 molecule and nothing else in the second equation?

Jessica Lutz 2E
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

Re: Question 8.55

Postby Jessica Lutz 2E » Thu Jan 11, 2018 5:46 pm

I think the 3/2 was added in the second equation because it wasn't balanced initially. In the equation given in the problem, there are 2 oxygens on the reactant side and 3 on the product side. When you add the 3/2 to just the oxygen on the reactant side, the equation is balanced.


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 2 guests