How do you find A in Arrhenius equation?

Mar 21, 2016

Plot a graph of $\frac{1}{T}$ versus $\ln \left(k\right)$. The $y$-intercept is $\ln \left(A\right)$.

Explanation:

The Arrhenius equation is

$k = A {e}^{- {E}_{\text{a}} / \left(R T\right)}$,

where $A$ and ${E}_{\text{a}}$ are assumed to be independent on the temperature.

This can be rewritten as

$\ln \left(k\right) = \ln \left(A\right) - {E}_{\text{a}} / R \cdot \frac{1}{T}$

If you measured the rate constant, $k$, at 2 different temperatures, ${T}_{1}$ and ${T}_{2}$, you can write

$\ln \left({k}_{1}\right) = \ln \left(A\right) - {E}_{\text{a}} / R \cdot \frac{1}{T} _ 1$
$\ln \left({k}_{2}\right) = \ln \left(A\right) - {E}_{\text{a}} / R \cdot \frac{1}{T} _ 2$

Solve them simultaneously and you will get

${E}_{\text{a}} = \frac{R \ln \left({k}_{2} / {k}_{1}\right)}{\frac{1}{T} _ 1 - \frac{1}{T} _ 2}$

$A = {e}^{\frac{{T}_{2} \ln \left({k}_{2}\right) - {T}_{1} \ln \left({k}_{1}\right)}{{T}_{2} - {T}_{1}}}$

If you measured the rate constant at many different temperatures, you can plot a graph of $\frac{1}{T}$ versus $\ln \left(k\right)$. If the Arrhenius relation is true, then the graph would look linear.

Extrapolate and get the $y$-intercept of the graph. That would be $\ln \left(A\right)$. You can get $A$ using $A = {e}^{\ln \left(A\right)}$.