How do you calculate the pre-exponential factor from the Arrhenius equation?

Apr 4, 2016

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 of $\frac{1}{{\text{M}}^{1 - m - n} \cdot s}$, where $m$ and $n$ are the order of reactant $A$ and $B$ in the reaction, respectively.
• $A$ is the pre-exponential factor, correlating with the number of properly-oriented collisions.
• ${E}_{a}$ is the activation energy in, say, $\text{J}$.
• $R$ is the universal gas constant, $\text{8.314472 J/mol"cdot"K}$.
• $T$ is the temperature in $\text{K}$.

So by taking the natural log, we can solve for $A$.

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

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

Thus, we have

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

Therefore, simply graph multiple values of $\ln k$ against their corresponding reciprocal temperatures, and you will get $\ln A$ as the y-intercept. Then just exponentiate $\ln A$.

$k$ can be gotten experimentally; just know your initial concentrations and reactant orders, and time the reaction accurately. Then determine the rate based on a particular reactant ($- \frac{d \left[A\right]}{\mathrm{dt}} = r \left(t\right)$) and solve for $k$ in the rate law.

However, ${E}_{a}$ is specific to the reaction and can't be obtained by observation, and neither can $A$. So this way of plotting $\ln k$ vs. $\frac{1}{T}$ makes it experimentally possible to determine $A$.

The less foolproof way that doesn't require multiple data points is to simply divide.

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

For this you would have to know the activation energy, rate constant, and temperature ahead of time, which you normally don't. Normally $k$, $T$, and $R$ are the only things you know.