Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Chemical kinetics diagramm -> chemical equation

  1. Mar 7, 2012 #1
    Hi everyone

    1. The problem statement, all variables and given/known data

    I got the three different diagramms (attached below) and I shall find a general chemical equation for all of them.

    2. Relevant equations


    3. The attempt at a solution

    I think I know how to do the third one, but I have troubles with the first and second one, because I have four different reactants. My thoughts for the third one are the following:

    [tex] blue \xrightarrow{k_{1}} red \xrightarrow{k_{2}} yellow[/tex]

    whereas k2 is far bigger than k1 because the intermediates curve is low, which says that it reacts fast.

    Can anyone approve this? Furthermore I need some hints for the first and second one. I'm absoulety not sure if I have something like red+yellow ->green or something like red->yellow->green

    Thank you for your help in advance.

    Attached Files:

  2. jcsd
  3. Mar 7, 2012 #2


    User Avatar

    Staff: Mentor

    Looks to me like in the first case green is still produced long after all red disappeared, so b->r->y->g looks the most logical. I am not so sure about the second. Could be r<->y is easily reversible.
  4. Mar 9, 2012 #3


    User Avatar
    Homework Helper
    Gold Member

    The first and the last in the series are evident I hope, so the only problem is the middle two.

    I can't see any way the second substance in the reaction sequence wouldn't peak before the third, can you?

    For Borek's suggestion of reversibility, if by that he means the second an third substance are in rapid equilibrium then they would peak at the same time, in fact the two curves would be identical when one is multiplied by the right factor.
  5. Mar 11, 2012 #4
    Thanks to both of you, I think I understood it now a bit better.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook