Question

The activation energy for a reaction at a temperature $ T_k $ was found to be $ 2.303RTmo{l^{ - 1}} $ the ratio of the rate constant to the Arrhenius factor is:
(A) $ {10^{ - 1}} $
(B) $ {10^{ - 2}} $
(C) $ 2 \times {10^{ - 3}} $
(D) $ 2 \times {10^{ - 2}} $

Answer 1

Hint: Here, in this equation we will use Arrhenius equation:
 $ k = A{e^{ - \dfrac{{E_a}}{{RT}}}} $
Where, $ k $ stands for the rate constant
 $ A $ stands for pre-exponential factor
 $ {E_a} $ stands for activation energy
 $ R $ stands for universal gas constant
 $ T $ stands for absolute temperature in Kelvin.
Now, with the help of this equation, we will further solve the question.

Complete answer:
Before solving this question, let’s understand first, what is the function of the Arrhenius equation. Arrhenius equation, basically, represents the fraction of collisions having enough energy which is needed to overcome the activation energy barrier at a particular temperature, T.
Now, let’s solve this question:
So for that, we will use the above mentioned equation known as Arrhenius equation:
 $ \Rightarrow $ $ k = A{e^{ - \dfrac{{E_a}}{{RT}}}} $
Let’s convert this equation into the another form, which will make it easier for us to solve then:
 $ \Rightarrow $ $ k = A{e^{ - \dfrac{{2.303RT}}{{RT}}}} $
In this equation, $ RT $ will get cancelled out, which will give us:
 $ \Rightarrow $ $ k = A{e^{ - 2.303}} $
This Arrhenius constant will be shifted from right hand side to the left hand side and then we will get:
 $ \Rightarrow $ $ \dfrac{k}{A} = {e^{ - 2.303}} $
And then we will solve it further and we will get:
 $ \Rightarrow $ $ \dfrac{k}{A} = {10^{ - 1}} $
Therefore, the correct option for this question is Option A- $ {10^{ - 1}} $ .

Note:
As we can see from the formula, the Arrhenius equation is a temperature dependent equation. This equation, as we have seen above, is used to determine the activation energy and also helps in determining the rate of the chemical reactions. We can also see from the equation that the rate constant exponentially increases with the decrease in activation energy.