# 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 thepre-exponential factor.#E_a# is theactivation energyin#"kJ/mol"# .#R# is theuniversal gas constant. We will be using#8.314472xx10^(-3)# #"kJ/mol"cdot"K"# .#T# is thetemperaturein#"K"# .

Now suppose we had *two* activation energies *rate constants* *same* reaction at the *same* temperature.

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 **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

#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.*