Question

How much faster is a reaction whose activation energy is `"52.3 kJ/mol"` compared to one at `"62.3 kJ/mol"`, both at `37^@ "C"`?

Answer 1

Recall the Arrhenius equation, which is:

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

where:

• `A` is the pre-exponential factor.
• `E_a` is the activation energy in `"kJ/mol"`.
• `R` is the universal gas constant. We will be using `8.314472xx10^(-3)` `"kJ/mol"cdot"K"`.
• `T` is the temperature in `"K"`.

Now suppose we had two activation energies `E_(a1)` and `E_(a2)` and respective rate constants `k_1` and `k_2`, but for the same reaction at the same temperature.

Then, the only things that would change are `k` and `E_a`:

`k_2 = Ae^(-E_(a2)"/RT")`
`k_1 = Ae^(-E_(a1)"/RT")`

Now, let's take the ratio of these to determine the "factor" by which the reaction rate is changed.

`(k_2 = Ae^(-E_(a2)"/RT"))/(k_1 = Ae^(-E_(a1)"/RT"))`

`color(green)((k_2)/(k_1)) = (e^(-E_(a2)"/RT"))/(e^(-E_(a1)"/RT"))`

Using the properties of exponents, we get:

`= e^(-E_(a2)"/RT" + E_(a1)"/RT")`

`= e^(-(E_(a2) - E_(a1))"/RT")`

`= color(green)(e^((E_(a1) - E_(a2))"/RT")`

Because we are solving for the ratio of the rates of reaction, we have to also relate `k` back to the rate law of the reaction to get:

`r_2(t) = k_2["reactant"]^"order"`

`r_1(t) = k_1["reactant"]^"order"`

Since we are looking at `k_2/k_1`, we don't really care what the reaction order is; it'll cancel out. Comparing these reactions we get:

`color(green)((r_2(t))/(r_1(t)) = k_2/k_1 = e^((E_(a1) - E_(a2))"/RT"`

We know that `E_(a1) = "62.3 kJ/mol"` and `E_(a2) = "52.3 kJ/mol"` at `T = "310.15 K"`. Therefore:

`color(blue)((r_2(t))/(r_1(t))) = e^(-(52.3 - 62.3 "kJ/mol")"/"(8.314472xx10^(-3) "kJ/mol"cdot"K"cdot"310.15 K")`

`~~ color(blue)(48.32)`

That means the new, catalyzed reaction is about 48 times as fast as the regular reaction.