Question

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.

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:

`log  k_2/k_1 = E_a/(2.303R) [(T_2-T_1)/(T_1T_2)]`

We get

`log  "2k"/k = E_a/(2.303xx8.314)[10/(298xx308)]`

`=>log 2 = E_a/(2.303xx8.314)[10/(298xx308)]`

`=> (E_a = 2.303xx8.314xx298xx308xxlog2)/10`

= 52897.78 J mol⁻¹

= 52.9 kJ mol⁻¹