15.3

Moderators: Chem_Mod, Chem_Admin

Adrian C 1D
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

15.3

Postby Adrian C 1D » Sun Mar 03, 2019 2:31 pm

In 20. s, the concentration of nitrogen dioxide, NO2, decreases from 450 (mmol)(L) to 320 (mmol)(L) in the reaction 2 NO2(g) => 2 NO(g) + O2(g). (a) Determine the rate of reaction of NO2. (b) Determine the rate of formation of O2.
(c) What is the unique rate of the reaction?

I'm confused about how to find c, should I use the rate law?

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

Re: 15.3

Postby Tam To 1B » Mon Mar 04, 2019 10:32 am

You divide what you got in part (a) by 2 because there are 2 mols of NO2 in the reaction. Another way is you can also divide (b) by one because there is one mol of O2 formed.
You divide the rate of reaction by the stoichiometric coefficient to get 3.3e-3 mol/Ls.

If you have aA --> bB + cC, the unique rate would be :
-1/a(d[A]/dt) = 1/b(d[B]/dt) = 1/c(d[C]/dt)


Return to “Method of Initial Rates (To Determine n and k)”

Who is online

Users browsing this forum: No registered users and 1 guest