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15 March, 18:58

A given reaction has an activation energy of 24.52 kj/mol. at 25°c, the half-life is 4 minutes. at what temperature will the half-life be reduced to 20 seconds?

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  1. 15 March, 20:37
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    According to half life equation:

    T (1/2) = ㏑2 / K1

    when the T (1/2) = 4 min * 60 = 240 sec

    by substitution:

    240 = 0.6931 / K1

    K1 = 2.9 x 10^-3

    when the second T (1/2) = 20 sec, so to get K2:

    T (1/2) = 0.6931 / K2

    by substitution:

    20 = 0.6931 / K2

    ∴K2 = 3.4 x 10^-2

    so, we can get T2 by using this formula:

    ㏑ (K2/K1) = Ea/R (1/T1 - 1/T2)

    by substitution:

    ㏑ (3.4 x 10^-2) / (2.9 x 10^-3) = (24520 / 8.314) (1/298 - 1/T2)

    ∴ T2 = 396.7 K

    = 396.7 - 273 = 123.7 °C
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