Question

A chemist's rule of thumb is that `10^@ "C"` of difference results in a two-fold change in the reaction rate. If the temperature is varied from `35^@ "C"` to `45^@ "C"` for the rxn of crystal violet with hydroxide, by what factor does the rate multiply?

Answer 1

You can use the Arrhenius equation.

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

where `A` is the frequency factor, `E_a` is the activation energy, `R` is the universal gas constant, and `T` is the temperature in `"K"`.

For the same reaction at most temperatures and pressures, the activation energy is assumed to not change. It is certainly true that `R` and `A` don't change with temperature.

Thus, all that can vary are `k` and `T`. Recall that the `i`th trial of any reaction can be expressed as a rate law:

`r_i(t) = k_i[A]_i^m[B]_i^n`

When we assume the concentrations are held constant as well to observe the effects of temperature change, we thus have:

`(r_i(t))/(r_j(t)) = (k_icancel([A]_i^m[B]_i^n))/(k_jcancel([A]_i^m[B]_i^n))`

for trial `j ne i`. Thus, the initial rates of reaction are directly proportional to the rate constants.

As a result, we can therefore find the ratio of the rate constants, which is also the ratio of the initial rates.

`k_i/k_j = (cancel(A)e^(-E_a"/"RT_i))/(cancel(A)e^(-E_a"/"RT_j))`

`= e^(-E_a"/"RT_i)*e^(E_a"/"RT_j)`

`= e^(-E_a"/"RT_i + E_a"/"RT_j)`

`= e^(-E_a/R(1/(T_i) - 1/(T_j))`

`= "exp"[-E_a/R(1/(T_i) - 1/(T_j))]`

If we take `T_i = 45 + 273.15 = "318.15 K"` and `T_j = 35 + 273.15 = "308.15 K"`, we thus say that `k_i` is for the reaction at the higher temperature.

A typical experimental result for the activation energy for this reaction of `"CV"^(+)` with `"OH"^(-)` is on the order of `20 ~ 60` `"kJ/mol"`. Take `E_a ~~ "50 kJ/mol"`. We get:

`k_i/k_j = color(blue)((r_i(t))/(r_j(t)))`

`= "exp"[-"50000 J/mol"/("8.314472 J/mol"cdot"K")*(1/"318.15 K" - 1/"308.15 K")]`

`= color(blue)(1.847)`

So it's not a bad estimate. It's somewhat close to `2`. If we chose a more accurate `E_a` we may have gotten closer to `2`.