The reactant concentration in a first-order reaction was 7.10×10−2 M after 30.0 s and 6.00×10−3 M after 60.0 s . ?

What is the rate constant for this reaction? Please show the correct units if possible.

2 Answers

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  • 1 month ago
    Favorite Answer

    Method 1:

    Integrated form of first-order reaction:

    ln[A] = -kt + ln[A]ₒ

    [A]ₒ = 7.10 × 10⁻² M

    When t = (60.0 - 30.0)s = 30.0 s, [A] = 6.00 × 10⁻³ M

    ln(6.00 × 10⁻³) = -k (30 s) + ln(6.00 × 10⁻³)

    k (30.0 s) = ln(7.10 × 10⁻²) - ln(6.00 × 10⁻³)

    k = [ln(7.10 × 10⁻²) - ln(6.00 × 10⁻³)] / (30.0 s)

    Rate constant, k = 8.24 × 10⁻² s⁻¹

    ====

    Method 2:

    Let n be the number of half-lives from 30.0 s to 60.0 s.

    (6.00 × 10⁻³)/(7.10 × 10⁻²) = (1/2)ⁿ

    (1/2)ⁿ = 6/71

    log(1/2)ⁿ = log(6/71)

    n log(1/2) = log(6/71)

    n = log(6/71) / log(1/2) = 3.565

    Half-life, t½ = [(60 - 30) s] / 3.565 = 8.415 s

    Rate constant, k = ln(2) / t½ = ln(2) / (8.415 s) = 8.24 × 10⁻² s⁻¹

  • ?
    Lv 7
    1 month ago

    Here is a solution courtesy of the Chemteam

    Set the first concentration to be Ao and the second to be A. The time will be 30. seconds.

    ln A = -kt + ln Ao

    ln 6.00 x 10-3 = - (k) (30. s) + ln 7.10 x 10-2

    -5.11599581 = - (k) (30. s) + -2.645075402

    2.471 = (k) (30. s) 

    k =0.0823 s-1 (to three sig figs)

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