The rate of a particular reaction doubles when temperature changes from 270 C to 370 C. Calculate the energy of activation of the reaction.

We are given that:
When T1 = 27 + 273 = 300 K
Let k1 = k
When T2 = 37 + 273 = 310 K
k2 = 2 k
Substituting these values the equation:
Log (k2 / k1)
= Ea / 2.303 R {(T2 – T1) / T1 T2}

We will get:
Log (2 k / k) = Ea / 2.303 x 8.314 {(310 – 300) / 300 x 310}
Log 2 = Ea / 2.303 x 8.314 x (10 / 300 x 310)
Ea = 53598.6 J mol-1
Ea = 53.6 kJ mol-1
Hence, the energy of activation of the reaction is 53.6 kJ mol-1

Category: Chemical Kinetics

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