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
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 using8.314472xx10^(-3) "kJ/mol"cdot"K" .T is the temperature in"K" .
Now suppose we had two activation energies
Then, the only things that would change are
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
r_2(t) = k_2["reactant"]^"order"
r_1(t) = k_1["reactant"]^"order"
Since we are looking at
color(green)((r_2(t))/(r_1(t)) = k_2/k_1 = e^((E_(a1) - E_(a2))"/RT"
We know that
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.