The rate of a reaction doubles when its temperature is raised from 298 K to 308 K. What is the activation energy of the reaction? (R = 8.314 J K−1 mol−1)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 52.9 kJ mol−1
The Arrhenius equation written for two temperatures relates the rate constants to the activation energy Ea. The rate doubles, so k2/k1 = 2, with T1 = 298 K and T2 = 308 K.
\( \displaystyle \log\frac{k_2}{k_1}=\frac{E_a}{2.303\,R}\left(\frac{T_2-T_1}{T_1T_2}\right) \)
\( \displaystyle \log 2=\frac{E_a}{2.303\times 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}}\left(\frac{10\ \mathrm{K}}{298\ \mathrm{K}\times 308\ \mathrm{K}}\right) \)
\( \displaystyle E_a=\frac{0.3010\times 2.303\times 8.314\times 91784}{10}\ \mathrm{J\,mol^{-1}}=52898\ \mathrm{J\,mol^{-1}} \)
\( \displaystyle E_a\approx 52.9\ \mathrm{kJ\,mol^{-1}} \)