By mathematically linking the Arrhenius equations of a reaction operating cleanly at two distinct te — Chemical Kinetics Chemistry Question
Question
By mathematically linking the Arrhenius equations of a reaction operating cleanly at two distinct temperatures ($T_1$ and $T_2$), the overall activation energy ($E_a$) can be isolated. What is the rigorously correct formulation of this dual-temperature equation using natural logarithms?
💡 Solution & Explanation
Integrating the Arrhenius formula between boundary states $T_1$ and $T_2$ yields $\ln k_2 - \ln k_1 = -\frac{E_a}{R}(\frac{1}{T_2} - \frac{1}{T_1})$. Factoring out the negative sign inside the parenthesis generates the correct bracket featuring the higher temperature $T_2$ positively leading the numerator: $\frac{E_a}{R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$.