If the initial concentration of CH3NC is 3.48×10-2 M, the concentration of CH3NC will be __  M after 577 s have passed.?

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  • 1 month ago
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    For a first order reaction, the integrated rate law is:

    ln [A] = -kt + ln [Ao]

    Substituting into this equation gives:

    ln[CH3NC] = -3.00X10^-3 /s (577 s) + ln (3.48X10^-2)

    [CH3NC] = 6.16X10^-3 M

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  • Dr W
    Lv 7
    1 month ago

    read through my answer here

    https://answers.yahoo.com/question/index?qid=20160...

    pay attention to this table

    .. .. .order.. .. .. . non-integrated.. .. .. . ..integrated

    .. .. . .. 0.. . .. . ...rate = k x [A]°... .. .. . . ..[At] = -kt + [Ao] 

    .. . .. .. 1.. . .. . ...rate = k x [A]¹... .. .. . . ln[At] = -kt + ln[Ao] 

    .. . .. . .2... . .. .. .rate = k x [A]²... .. .. . . 1/[At] = +kt + 1/[Ao]

    note the difference between integrated and non-integrated

    .. non-integrated.... relates RATE with CONCENTRATION

    .. .. .. .integrated.. ..relates TIME with CONCENTRATION

    ***********

    your problem is "time" vs "concentration" and is 1st order so we use

    .. ln[At] = -kt + ln[Ao]

    rearranging

    .. ln( [At] / [Ao] ) = -kt

    .. [At] / [Ao] = exp(-kt)

    .. [At] = [Ao] * exp(-kt)

    solving

    .. [At] = 3.48x10^-2M * exp(-3.00x10^-3/sec * 577sec)

    ... . ...= 6.16x10^-3M

    Source(s): t
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