2016 Final


Moderators: Chem_Mod, Chem_Admin

Blake_Katsev_2E
Posts: 113
Joined: Wed Sep 21, 2016 2:57 pm

2016 Final

Postby Blake_Katsev_2E » Thu Mar 16, 2017 5:42 pm

For student B in problem four, how would you write the rate law because there are two intermediates? do you need to cancel out the O2's on each side to make it only have one intermediate or is there another method?

Sangita_Sub_3H
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

Re: 2016 Final

Postby Sangita_Sub_3H » Thu Mar 16, 2017 7:58 pm

I think to write the overall rate law, the equation would be 2ClO(g)+ O2(g) -->2ClO2(g), in which the O2 from step one and the 2O2 from step two would be cancelled and become 1 O2(g). Therefore, the rate law would be rate = k[ClO2]2[O2].


Return to “Reaction Mechanisms, Reaction Profiles”

Who is online

Users browsing this forum: No registered users and 1 guest