According to standard Arrhenius kinetics, plotting on the y-axis against on the x-axis produces a st — Chemical Kinetics Chemistry Question
Question
According to standard Arrhenius kinetics, plotting $\log_{10} k$ on the y-axis against $1/T$ on the x-axis produces a straight line. If the experimental slope of this line evaluates to $-10^4 \text{ K}$, what is the absolute activation energy $E_a$ of this reaction? (Given $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$)
Answer: A
💡 Solution & Explanation
The base-10 logarithmic form of the Arrhenius equation is $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$. The slope of the resulting plot corresponds exactly to $m = -\frac{E_a}{2.303 R}$. We are provided slope $= -10^4$. Thus, $-\frac{E_a}{2.303 \times 8.314} = -10000 \Rightarrow E_a = 10000 \times 2.303 \times 8.314 = 191471 \text{ J mol}^{-1} \approx 191.4 \text{ kJ mol}^{-1}$.
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