Qing 發問於 科學及數學化學 · 9 年前

order of reaction!

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  • 9 年前
    最愛解答

    b. (i)

    Let Rate = k [I2]^a [CH3COCH3]^b [H^+]^c

    3.5 x 10^-5 = k (2.5 x 10^-4)^a (2.0 x 10^-1)^b (5.0 x 10^-3)^c ...... (1)

    3.5 x 10^-5 = k (1.5 x 10^-4)^a (2.0 x 10^-1)^b (5.0 x 10^-3)^c ...... (2)

    1.4 x 10^-4 = k (2.5 x 10^-4)^a (4.0 x 10^-1)^b (1.0 x 10^-2)^c ...... (3)

    7.0 x 10^-5 = k (2.5 x 10^-4)^a (4.0 x 10^-1)^b (5.0 x 10^-3)^c ...... (4)

    (1)/(2):

    1 = (5/3)^a

    a = 0

    (3)/(4):

    2 = 2^c

    c = 1

    (1)/(4):

    (1/2) = (1/2)^b

    b = 1

    Hence, Rate = k [CH3COCH3][H^+]

    b. (ii)

    Put the first set of data into the rate equation :

    3.5 x 10^-5 = k (2.0 x 10^-1) (5.0 x 10^-3)

    Rate constant, k = 3.5 x 10^-2 (mol dm^-3)^-1 s^-1

    (c)

    When pH = 4 :

    [H^+] = 10^-pH = 1 x 10^-4 M

    Rate = (3.5 x 10^-2) x [CH3COCH3] x (1 x 10^-4)

    Hence, Rate = (3.5 x 10^-6) x [CH3­COCH3]

    It becomes a pseudo-first-order reaction in the form :

    Rate = k' [CH3COCH3]

    Half life of the pseudo-first-order reaction, t1/2

    = ln2/k'

    = ln2/(3.5 x 10^-6)

    = 1.98x 10^5 s

    資料來源: micatkie
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