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

### 2 Answers

- hcbiochemLv 71 month agoFavorite Answer
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 WLv 71 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

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