When the temperature of a reaction increases from 27°C to 37°C, the rate increases by 2.5 times, the activation energy in the temperature range is:

70.8 kJ

The correct answer is option 1. 70.8 kJ.

Given,

\(\frac{k_2}{k_1} = 2.5\)

\(T_1 = 300 K\)

\(T_2 = 310 K\)

According to Arrhenius equation,

\(log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R}\left[\frac{T_2 − T_1}{T_1T_2}\right]\)

\(⇒ log(2.5) = \frac{E_a}{2.303 × 8.314}\left[\frac{310 − 300}{300 × 310}\right]\)

\(⇒ E_a = 70.77 × 10^{3} J\)

\(⇒ E_a = 70.8 kJ\)