🤓 Based on our data, we think this question is relevant for Professor Bindell's class at UCF.

$\overline{){\mathbf{ln}}{\mathbf{}}{\mathbf{k}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}}{\mathbf{}}\left(\frac{\mathbf{1}}{\mathbf{T}}\right){\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{ln}}{\mathbf{}}{\mathbf{A}}}\phantom{\rule{0ex}{0ex}}{\mathbf{y}}{\mathbf{}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{m}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}\left(\mathbf{x}\right){\mathbf{}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{b}}$

We are given the slope (m) = -8500 K^{-1}. We shall compute activation energy (E_{a}) from this slope:

The rate constant of a reaction is measured at different temperatures. A plot of the natural log of the rate constant as a function of the inverse of the temperature (in kelvins) yields a straight line with a slope of -8500 K^{-1} . What is the activation energy (E_{a}) for the reaction?

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Based on our data, we think this problem is relevant for Professor Bindell's class at UCF.