Question

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

Answer 1

By graphing.

The Arrhenius equation is

`\mathbf(k = Ae^(-E_a"/"RT))`,

where:

• `k` is the rate constant, in units of `1/("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, `"J"`.
• `R` is the universal gas constant, `"8.314472 J/mol"cdot"K"`.
• `T` is the temperature in `"K"`.

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

`ln k = lnAe^(-(E_a)/(RT))`

`= lnA - (E_a)/(RT)`

Thus, we have

`color(blue)(stackrel(y)overbrace(ln k) = stackrel(m)overbrace(-(E_a)/(R))stackrel(x)overbrace(1/T) + stackrel(b)overbrace(lnA))`

Therefore, simply graph multiple values of `lnk` against their corresponding reciprocal temperatures, and you will get `lnA` as the y-intercept. Then just exponentiate `lnA`.

`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 (`-(d[A])/(dt) = r(t)`) 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 `lnk` vs. `1/T` makes it experimentally possible to determine `A`.

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

`color(green)(A = k/(e^(-E_a"/"RT)))`

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.