the rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298 K calculate Ea.​

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1. Answer:

   It is given that T 1= 298 K

   ∴T = (298 + 10) K

   = 308 K

   We also know that the rate of reaction doubles when the temperature is increased by 10 K.

   Therefore, let us take the value of k1 = k and that of k2 = 2k

   Also, R = 8.314 J K – 1 mol – 1

   Now, substituting these values in the equation:

   log k2/k1=Ea/2.303*R(1/T1-1/T2)

   Ea=52.9kJ mol

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