Question

According to Arrhenius equation, the slope of $\log \,k \to \dfrac{1}{T}$ plot is:
A.$\dfrac{{ - {E_a}}}{{2.303R}}$
B.$\dfrac{{ - {E_a}}}{{2.303}}$
C.$\dfrac{{ - {E_a}}}{{2.303RT}}$
D.$\dfrac{{{E_a}}}{{2.303RT}}$

Answer 1

Hint: To answer this question, you should recall the concept of the Arrhenius equation and plotting graphs. Extrapolate the Arrhenius equation using log function to achieve a form in terms of the general equation of a straight line. Compare and plot the graph to find the answer to this question.
Formula used:
The Arrhenius equation
 \[k = A{e^{ - {E_a}/RT}}\],
where $A$: The frequency or pre-exponential factor, $T$: temperature, \[{E_a}\] : Activation energy and $R$: Universal gas constant

Complete step by step answer:
We know that \[{e^{ - {E_a}/RT}}\] in the Arrhenius equation is the fraction of collisions that have enough energy to react i.e., have energy greater than or equal to the activation energy \[{E_a}\]. This equation is used to study the dependence of a reaction on temperature.
As we know, the Arrhenius equation is given as follows:
\[k = A{e^{ - {E_a}/RT}}\], taking log on both sides.
\[{\text{ln}}k = {\text{ln}}A - {E_a}/RT\].
Converting this equation:
\[2.303\log k = 2.303\log A - {E_a}/RT\]
Now compare this with the general equation of a straight line $y = mx + c$.
The slope of $\ln k$v/s $\dfrac{1}{T}$can be written as:
$m = $$\dfrac{{ - {E_a}}}{{2.303R}}$

Therefore, the correct option is option A.

Note:
The Arrhenius equation is not only simple but a remarkably accurate formula too. Not only it is used to study reaction rates but also to model the temperature variance of permeation, diffusion and solubility coefficients, and other chemical processes over moderate temperature ranges. You should know the importance of the Arrhenius equation. $RT$ is the average kinetic energy, and the exponent is just the ratio of the activation energy \[{E_a}\] to the average kinetic energy. Larger this ratio, the smaller the rate. It can be concluded that high temperature and low activation energy favour larger rate constants, and thus speed up the reaction.