Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and, as of 20 April 2021 (Eastern Time), the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
I know this is second order so would it be t1/2= 1/K[A]0 but how would i implement the time and concentration.
I thought it would it be t1/2= 1/0.806*[.154]*(31.7) but that doesnt make sense because its asking for the concentration so do we need to use the second order integrated law? 1/[A]t= kt +1/[A]0 which would be 1/[A]t= 0.806*31.7s + 1/[.154]
That is my best guess, but still wrong.
Any help appreciated.
- Dr WLv 71 month ago
from the integrated 2nd order rate equation
.. 1/[At] = +kt + 1/[Ao]
.. [At] = the concentration of A after a time of t has elapsed.
.. .. . . . .i.e.. [A] at time = t.... not [A] times t.
.. . .. ..it's the concentration of t when 31.7 seconds have ticked off the clock.
.. k = rate constant
.. t = elapsed time
.. [Ao] = initial concentration of A.. at the start fo the experiment.. when t = 0
.. 1/[A] = (0.806 M-1 s-1) * 31.7s + 1/(0.154) M-1
.. [A] = what?
do the math
.. 0.806 * 31.7 + 1/0.154 = what
then take the reciprocal.