The rate constants of a reaction at 500 K and 700 K are 0.02 s^-1 and 0.07 s^-1 respectively. What is the activation energy? (log(0.07/0.02) = 0.544, R = 8.314 J K^-1 mol^-1)

1. $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left[\dfrac{T_2 - T_1}{T_1 T_2}\right]$. 2. $\dfrac{T_2-T_1}{T_1T_2} = \dfrac{700-500}{700\times500} = \dfrac{200}{350000} = 5.714\times10^{-4}$. 3. $0.544 = \dfrac{E_a\times5.714\times10^{-4}}{2.303\times8.314} = \dfrac{E_a\times5.714\times10^{-4}}{19.15}$. 4. $E_a = \dfrac{0.544\times19.15}{5.714\times10^{-4}}$. 5. $= 18231\ \text{J mol}^{-1}$. 6. Distractor C is correct; distractor B doubles it by omitting the 2.303R divisor properly. _Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.20_