How do you solve the Arrhenius equation for activation energy?

Apr 1, 2016

Generally, it can be done by graphing. The Arrhenius equation is:

$\setminus m a t h b f \left(k = A {e}^{- {E}_{a} \text{/} R T}\right)$

where:

• $k$ is the rate constant, in units that depend on the rate law. For instance, if $r \left(t\right) = k {\left[A\right]}^{2}$, then $k$ has units of "M"/"s" * 1/"M"^2 = 1/("M"cdot"s").
• $A$ is the "pre-exponential factor", which is merely an experimentally-determined constant correlating with the frequency of properly oriented collisions.
• ${E}_{a}$ is the activation energy in units of, say, $\text{J}$.
• $R$ is the universal gas constant. Using $\text{J}$, $R = \text{8.314472 J/mol"cdot"K}$.
• $T$ is temperature in $\text{K}$.

METHOD A

To "solve for it", just divide by $A$ and take the natural log.

$\frac{k}{A} = {e}^{- {E}_{a} \text{/} R T}$

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

$\ln k - \ln A = - \frac{{E}_{a}}{R T}$

$\textcolor{b l u e}{\stackrel{y}{\overbrace{\ln k}} = \stackrel{m}{\overbrace{- \frac{{E}_{a}}{R}}} \stackrel{x}{\overbrace{\frac{1}{T}}} + \stackrel{b}{\overbrace{\ln A}}}$

So now, if you grab a bunch of rate constants for the same reaction at different temperatures, graphing $\ln k$ vs. $\frac{1}{T}$ would give you a straight line with a negative slope.

The slope is $m = - \frac{{E}_{a}}{R}$, so now you can solve for ${E}_{a}$. You can also easily get $A$ from the y-intercept.

METHOD B

Or, if you meant literally solve for it, you would get:

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

$\textcolor{b l u e}{{E}_{a} = - R T \ln \left(\frac{k}{A}\right)}$

So knowing the temperature, rate constant, and $A$, you can solve for ${E}_{a}$.

However, since $A$ is experimentally determined, you shouldn't anticipate knowing $A$ ahead of time (unless the reaction has been done before), so the first method is more foolproof. Furthermore, using $k$ and $T$ for one trial is not very good science. It's better to do multiple trials and be more sure.