An Arrhenius graphical plot of versus (where is in Kelvin) for a chemical reaction rigorously yields — Chemical Kinetics Chemistry Question
Question
An Arrhenius graphical plot of $\ln k$ versus $1/T$ (where $T$ is in Kelvin) for a chemical reaction rigorously yields a straight descending line with a calculated mathematical slope of exactly $-4606 \text{ K}$. Given that the universal gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, the activation energy ($E_a$) of this specific reaction evaluates to:
💡 Solution & Explanation
The natural logarithm Arrhenius equation is $\ln k = \ln A - \frac{E_a}{RT}$. When plotting $\ln k$ against $1/T$, the resulting straight-line slope $m$ identically equals $-\frac{E_a}{R}$. We are given slope $= -4606$. Thus, $-\frac{E_a}{R} = -4606 \Rightarrow E_a = 4606 \times R = 4606 \times 8.314 = 38294.28 \text{ J mol}^{-1}$. Converting to kilojoules gives $\approx 38.29 \text{ kJ mol}^{-1}$.