The rate of a chemical reaction is observed to exactly double for a temperature rise strictly from t — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

The dual-temperature Arrhenius equation states $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$. Here, $\frac{k_2}{k_1} = 2$, $T_1 = 298$, $T_2 = 308$, and $\Delta T = 10$. $\ln 2 = \frac{E_a \times 10}{R \times 298 \times 308} \Rightarrow E_a = \frac{R \times 298 \times 308 \times \ln 2}{10}$. Since $\ln 2 = 2.303 \log 2 \approx 0.693$, expressions A, B, and C are all algebraically identical derivations.