When the temperature of a reaction increases from 27°C to 37°C, the rate increases 2.5 times. The activation energy in this temperature range is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 70.8 kJ mol−1
The Arrhenius equation links the rate constants at two temperatures to the activation energy Ea:
\( \displaystyle \log\frac{k_2}{k_1}=\frac{E_a}{2.303\,R}\left(\frac{T_2-T_1}{T_1T_2}\right) \)
At the same concentrations the rate is proportional to k, so the rate ratio equals the rate-constant ratio: \( \frac{k_2}{k_1}=2.5 \). The temperatures are T1 = 27 + 273 = 300 K and T2 = 37 + 273 = 310 K.
\( \displaystyle \log 2.5=\frac{E_a}{2.303\times8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}}\left(\frac{310-300}{300\times310}\right) \)
\( \displaystyle E_a=\frac{2.303\times8.314\times300\times310\times0.3979}{10} \)
\( \displaystyle E_a=70853\,\mathrm{J\,mol^{-1}}\approx70.8\,\mathrm{kJ\,mol^{-1}} \)