It is given that T₁ = 298 K

∴ T₁ = (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 \(\frac{k_2}{k_1}\) = \(\frac{E_a}{2.303 R}\)\(\Big[\frac{T_2 -T_1}{T_1T2}\Big]\)

We get,

\(\frac{2k}{k}\) = \(\frac{E_a}{2.303 \times 8.314}\)\(\Big[\frac{10}{298 \times 308}\Big]\)

⇒ log 2 = \(\frac{E_a}{2.303 \times 8.314}\)\(\Big[\frac{10}{298 \times 308}\Big]\)

⇒ E_a = \(\frac{2.303 \times 8.314 \times 298 \times 308 \times log 2}{10}\)

= 52897.78 J mol^-1

= 52.9 kJ mol^-1