CHEMISTRY TUTORS PLEASE HELP!?

A first order reaction has a rate law

of K= 8.23 x 10-2 1/sec .

If the reaction starts with a concentration of A [A]=3.24

M and is stopped at [A] = 0.79 M, how much time has

passed?

3 Answers

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  • 1 month ago

    First order kinetics is described by the integrated rate law:

    [A] = [A]o e^(-kt) ...... where t is the elapsed time and k is the rate constant.

    The equation can also be written a:

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

    ln[A] - ln[A]o = -kt

    t = ln([A] / [A]o) / -k

    t = ln(0.79 / 3.24) / -8.23x10^-2 s⁻¹

    t = 17.1 s

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  • Dr W
    Lv 7
    1 month ago

    I cover the derivation of rate equations in my answer here

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

    note this table... (and MEMORIZE IT!!!)

    .. .. .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 non-integrated and integrated

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

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

    *********

    This problem of yours states

    .. first order

    .. concentration

    .. time

    so we use this equation

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

    *************

    rearranging that 1st order integrated rate equation

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

    .. kt = ln([Ao]/[At])...  recall that ln(a) - ln(b) = ln(a/b)

    .. t = ln([Ao]/[At]) / k

    solving

    .. t = ln(3.24M / 0.79M) / (8.23x10^-2/s) = 17.1 sec

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  • 1 month ago

    Since the reaction is first order, it has a half life (t 1/2) = 0.693/k

    So, t 1/2 = 0.693/8.23x10^-2/sec

    t 1/2 = 8.42 sec

    Next, we can use a simple expression for the first order decay:

    fraction remaining = 0.5^n where n = the number of half lives elapsed

    In this problem we are given (or can find) the fraction remaining....

    0.79 M / 3.24 M = 0.2438 = fraction remaining 

    Next, using this value, we can find how many half lives have elapsed....

    fraction remaining = 0.5^n

    0.2438 = 0.5^n

    log 0.2438 = n log 0.5

    n = 2.036 half lives

    Finally, we simply convert 2.036 half lives to seconds, minutes...whatever.....

    2.036 half lives x 8.42 sec/half life = 17.1 seconds has passed

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