Question

How do you find the A and ${{\text{E}}_{\text{a}}}$ using the Arrhenius equation?

Answer 1

Hint: To answer this question we should know what the Arrhenius equation is. The equation relates to the rate constant, pre-exponential factor, activation energy, and the temperature is known as the Arrhenius equation. The name of the equation is based upon the name of the scantiest who gave this equation. We will rearrange the Arrhenius equation for the desired value.

Complete step-by-step answer:The relation in the rate constant, pre-exponential factor, activation energy, and the temperature is as follows:
${\text{K}}\,{\text{ = }}\,{\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}$
Where,
K is the rate constant
A is the pre-exponential factor
${{\text{E}}_{\text{a}}}$ is the activation energy
R is the gas constant
T is the temperature
The above equation is known as the Arrhenius equation.
We have to determine the A and ${{\text{E}}_{\text{a}}}$ using the Arrhenius equation. Let’s determine the Ea first. We rearrange the Arrhenius equation for ${{\text{E}}_{\text{a}}}$ as follows:
${\text{K}}\,{\text{ = }}\,{\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}$
We will take natural log on both sides of the equation as follows;
${\text{ln}}\,{\text{K}}\,{\text{ = }}\,\ln \,{\text{A}} - \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}$….$(1)$
On rearranging the equation for ${{\text{E}}_{\text{a}}}$ we get,
$\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}\, = \,\,{\text{ln}}\,{\text{K}} - \,\ln \,{\text{A}}$
We know, $\ln {\text{a}} - \ln {\text{b}}\, = \,\ln \dfrac{{\text{a}}}{{\text{b}}}$ so,
$\Rightarrow \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}\, = \,\,{\text{ln}}\,\dfrac{{\text{K}}}{{\text{A}}}$
$\Rightarrow {{\text{E}}_{\text{a}}}\, = \,\,{\text{RT}}\,{\text{ln}}\,\dfrac{{\text{K}}}{{\text{A}}}$
So, the equation used for the calculation of ${{\text{E}}_{\text{a}}}$ is ${{\text{E}}_{\text{a}}}\, = \,\,{\text{RT}}\,{\text{ln}}\,\dfrac{{\text{K}}}{{\text{A}}}$.
Now we will rearrange the equation $(1)$for A as follows:
${\text{ln}}\,{\text{K}}\,{\text{ = }}\,\ln \,{\text{A}} - \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}$….$(1)$
$\Rightarrow {\text{ln}}\,{\text{A}}\,{\text{ = }}\,\ln \,{\text{K}} + \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}$
$\Rightarrow {\text{A}}\,{\text{ = }}\,{\text{K}}{{\text{e}}^{{{\text{E}}_{\text{a}}}{\text{/RT}}}}$
So, the equation used for the calculation of A is${\text{A}}\,{\text{ = }}\,{\text{K}}{{\text{e}}^{{{\text{E}}_{\text{a}}}{\text{/RT}}}}$.
Therefore, we can determine the A and ${{\text{E}}_{\text{a}}}$ by rearranging the Arrhenius equation for A and ${{\text{E}}_{\text{a}}}$.

Note: We can also compare the Arrhenius equation at two different temperatures for the calculation of activation energy. The equation obtained by comparing the Arrhenius equation at two different temperature is,
\[{\text{ln}}\,\dfrac{{{{\text{K}}_{\text{2}}}}}{{{{\text{K}}_{\text{1}}}}}\, = \, - \dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}}\left[ {\dfrac{1}{{{{\text{T}}_2}}} - \dfrac{{\text{1}}}{{{{\text{T}}_1}}}} \right]\]
We can also determine the A and ${{\text{E}}_{\text{a}}}$ from Arrhenius equation by plotting the graph between ln K and $1/{\text{T}}$.