Arrhenius Law (Lecture 5)

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

$$k=A{e}^{-E_a/RT}\text{\hspace{1em}(1)}$$

If we take the log of both sides:

$$\begin{array}{rl}\ln k& =\ln A+\ln e^{-E_a/RT}\\ \ln k& =\ln A-\frac{E_a}{RT}\end{array}\text{\hspace{1em}(2)}$$

we get an equation of the form $y=b+mx$ where $y=\ln k$, $b=\ln A$, $m=-\frac{E_a}{R}$, $x=1/T$. 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 $k=A{e}^{-E_a/RT}$, 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:

$$\begin{array}{rl}\ln k_1& =\ln A-\frac{E_a}{R{T}_1}\\ \ln k_2& =\ln A-\frac{E_a}{R{T}_2}\end{array}$$

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 $\ln A$ from both equations:

$$\begin{array}{rl}\ln k_1+\frac{E_a}{R{T}_1}& =\ln A\\ \ln k_2+\frac{E_a}{R{T}_2}& =\ln A\end{array}$$

If the right-hand side is equal, so should be left-hand-sides:

$$\begin{array}{rl}\ln k_1+\frac{E_a}{R{T}_1}& =\ln k_2+\frac{E_a}{R{T}_2}\\ \frac{E_a}{R{T}_1}-\frac{E_a}{R{T}_2}& =\ln k_2-\ln k_1\\ \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)& =\ln \frac{k_2}{k_1}\\ E_a=\frac{R\ln \frac{k_2}{k_1}}{\frac{1}{T_1}-\frac{1}{T_2}}& \end{array}$$

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 $k_2/k_1>1$, $\ln k_2/k_1>0$, 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.