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Anonymous asked in Science & MathematicsChemistry · 1 month ago

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. 

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1 Answer

  • Dr W
    Lv 7
    1 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.

    ..   . ..'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.

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