Agiven 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?

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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