Question

Calculate the energy of activation of a reaction for which rate constant becomes doubles by Increase of 10$K$ temperature from 298$K$

Answer 1

It is given that T1 = 298 K

∴T2 = (298 + 10) K

= 308 K

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

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:

= 52897.78 J mol - 1

= 52.9 kJ mol - 1