In this class, we mostly treat Arrhenius law as something given. If you’re curious about its derivation, visit notes on Collision Theory

$k=Ae_{−E_{a}/RT}(1)$If we take the log of both sides:

$lnklnk =lnA+lne_{−E_{a}/RT}=lnA−RTE_{a} (2) $we get an equation of the form $y=b+mx$ where $y=lnk$, $b=lnA$, $m=−RE_{a} $, $x=1/T$. See notes on straight line form to see reasoning as to why we might want to bring it to a linear form.

To reiterate, imagine you’re an experimentalist, and you calculate values of $k$ at different temperatures $T$. Can you infer $E_{A}$ and $A$? Yes, if you can fit an equation $k=Ae_{−E_{a}/RT}$, which you can’t do in Excel (and remain a sane person). However, once you convert it to the straight line form, you’ll find the values of parameters $b$ and $m$ from which you can find $A$ and $E_{a}$.

Also it makes it easy to do estimations from two points. Imagine you have $k_{1}$ and $k_{2}$ at temperatures $T_{1}$ and $T_{2}$. We start by writing eq(2) for both constants:

$lnk_{1}lnk_{2} =lnA−RT_{1}E_{a} =lnA−RT_{2}E_{a} $We have a system of 2 equations with 2 unknowns, so it should have a unique solution. One way to find it is to express $lnA$ from both equations:

$lnk_{1}+RT_{1}E_{a} lnk_{2}+RT_{2}E_{a} =lnA=lnA $If the right-hand side is equal, so should be left-hand-sides:

$lnk_{1}+RT_{1}E_{a} RT_{1}E_{a} −RT_{2}E_{a} RE_{a} (T_{1}1 −T_{2}1 )E_{a}=T_{1}1 −T_{2}1 Rlnk_{1}k_{2} =lnk_{2}+RT_{2}E_{a} =lnk_{2}−lnk_{1}=lnk_{1}k_{2} $In case you get confused by signs (Is it $1/T_{1}−1/T_{2}$ or $1/T_{2}−1/T_{1}$)? Remember that if $k$ increases with temperature, then $E_{a}$ has to be positive. With $k_{2}/k_{1}>1$, $lnk_{2}/k_{1}>0$, so the denominator has to be positive as well. Thus, we should subtract a smaller number from a larger. Because we deal with inverse temperatures, we should subtract 1/larger temperature from 1/smaller temperature.