Question 572811
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The difference of the logs is the log of the quotient.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-E_a}{R}\left(\frac{1}{T_1}\ -\ \frac{1}{T_2}\right)\ =\ \ln\left(\frac{k_2}{k_1}\right)]


Multiply both sides by *[tex \LARGE R] and divide by the negative of the difference between the temperature reciprocals:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ E_a\ =\ \frac{R\,\cdot\,\ln\left(\frac{k_2}{k_1}\right)}{\frac{1}{T_2}\ -\ \frac{1}{T_1}}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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