Question

How do you find A in Arrhenius equation?

Answer 1

Plot a graph of `1/T` versus `ln(k)`. The `y`-intercept is `ln(A)`.

Explanation:

The Arrhenius equation is

`k = A e^{-E_"a"/(RT)}`,

where `A` and `E_"a"` are assumed to be independent on the temperature.

This can be rewritten as

`ln(k) = ln(A) - E_"a"/R * 1/T`

If you measured the rate constant, `k`, at 2 different temperatures, `T_1` and `T_2`, you can write

`ln(k_1) = ln(A) - E_"a"/R * 1/T_1`
`ln(k_2) = ln(A) - E_"a"/R * 1/T_2`

Solve them simultaneously and you will get

`E_"a" = frac{Rln(k_2/k_1)}{1/T_1 - 1/T_2}`

`A = e^{frac{T_2ln(k_2) - T_1ln(k_1)}{T_2 - T_1}}`

If you measured the rate constant at many different temperatures, you can plot a graph of `1/T` versus `ln(k)`. 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(A)`. You can get `A` using `A = e^{ln(A)}`.