Question

Identify the correct relation between rate constant $(k)$, activation energy $({{\text{E}}_a})$, temperature T $({\text{K}})$?
A.$\log \dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{2.303{\text{R}}}}\left[ {\dfrac{1}{{{{\text{T}}_1}}} - \dfrac{1}{{{{\text{T}}_2}}}} \right]$
B.$\log \dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}}\left[ {\dfrac{1}{{{{\text{T}}_1}}} - \dfrac{1}{{{{\text{T}}_2}}}} \right]$
C.$\log \dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{ - {{\text{E}}_{\text{a}}}}}{{\text{R}}}\left[ {\dfrac{1}{{{{\text{T}}_1}}} - \dfrac{1}{{{{\text{T}}_2}}}} \right]$
D,$\log \dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{ - {{\text{E}}_{\text{a}}}}}{{2.303{\text{R}}}}\left[ {\dfrac{1}{{{{\text{T}}_1}}} - \dfrac{1}{{{{\text{T}}_2}}}} \right]$

Answer 1

To answer this question we must have the knowledge of the Arrhenius equation. Write the two Arrhenius equations for two different variables and rearrange the equations to get the desired relation between the three of them.

Complete step by step answer:
The expression for Arrhenius equation is ${\text{k}} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}/{\text{RT}}}}$
Here K is the rate constant, A is Arrhenius constant or pre- exponential factor or frequency factor, ${{\text{E}}_a}$ is activation energy, R is gas constant and T is temperature. According to Arrhenius the rate constant is a function of temperature, if we will change the temperature the value of the rate constant will also change but activation energy is not a function of temperature. If we change the temperature activation energy will remain constant. The above fact is based on the postulates of Arrhenius theory. Let the two temperature be ${{\text{T}}_1}$ and ${{\text{T}}_2}$ and ${{\text{k}}_1}$ and ${{\text{k}}_2}$ be the rate constant respectively. Hence we will get 2 equations as follow:
${{\text{k}}_1} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_1}}}$ and ${{\text{k}}_2} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_2}}}$
Dividing equation 2 with 1 we will get:
$\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{{\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_2}}}}}{{{\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_1}}}}}$
We will cancel the constant quantities and take the exponential power to numerator therefore its power will become positive:
$\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = {{\text{e}}^{ - {{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_2}}} \times {{\text{e}}^{{{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_1}}}$
Now we can add powers if the base is same, so adding the power of both the exponentials,
$\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = {{\text{e}}^{ - {{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_2} + {{\text{E}}_{\text{a}}}/{\text{R}}{{\text{T}}_1}}}$
We need to take natural log both sides to cancel exponential, because exponential and natural log are reciprocal to each other. Hence we will get:
${\text{ln}}\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{{\text{ - }}{{\text{E}}_{\text{a}}}}}{{{\text{R}}{{\text{T}}_{\text{1}}}}}{\text{ + }}\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{R}}{{\text{T}}_{\text{2}}}}}$
Rearranging we will get,
${\text{ln}}\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}}\left[ {\dfrac{1}{{{{\text{T}}_{\text{2}}}}} - \dfrac{{\text{1}}}{{{{\text{T}}_{\text{1}}}}}} \right]$
To convert natural log into log with base 10 we need to multiply natural log by $2.303$
${\text{2}}{\text{.303log}}\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}}\left[ {\dfrac{1}{{{{\text{T}}_2}}} - \dfrac{1}{{{{\text{T}}_1}}}} \right]$
Hence the required equation is
${\text{log}}\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303R}}}}\left[ {\dfrac{1}{{{{\text{T}}_{\text{2}}}}} - \dfrac{1}{{{{\text{T}}_{\text{1}}}}}} \right]$

The correct option is A.

Note:
However it was stated by Arrhenius that activation energy is independent of temperature, but this is not actually true. Activation energy varies linearly with temperature. Activation energy is the energy required by reactant so that they overcome the energy gap between reactant and product and product can be formed.