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"?

1 Answer
Mar 3, 2016

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.