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# Second order?

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.

### 1 Answer

- Dr WLv 71 month ago
from the integrated 2nd order rate equation

.. 1/[At] = +kt + 1/[Ao]

where

.. [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

so.

.. 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.