Question

How do you solve the Arrhenius equation for activation energy?

Answer 1

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

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

where:

• `k` is the rate constant, in units that depend on the rate law. For instance, if `r(t) = k[A]^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, `"J"`.
• `R` is the universal gas constant. Using `"J"`, `R = "8.314472 J/mol"cdot"K"`.
• `T` is temperature in `"K"`.

METHOD A

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

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

`ln(k/A) = -(E_a)/(RT)`

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

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

So now, if you grab a bunch of rate constants for the same reaction at different temperatures, graphing `lnk` vs. `1/T` would give you a straight line with a negative slope.

The slope is `m = -(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(k/A) = -(E_a)/(RT)`

`color(blue)(E_a = -RTln(k/A))`

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.