Question

A straight line with a negative slope was found in a graph of ${\text{lo}}{{\text{g}}_{\text{e}}}{\text{K vs }}{{\text{T}}_{\text{1}}}$ ​ with an angle of ${45^o}$ with x-axis. The energy of activation (in cal) is:
A.2
B.1
C.4
D.None of these

Answer 1

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: ${\text{k = A}}{{\text{e}}^{\dfrac{{ - {\text{Ea}}}}{{{\text{RT}}}}}}$, where ${\text{A}}$: The frequency or pre-exponential factor,
${\text{T}}$: temperature, ${{\text{E}}_{\text{a}}}$ : Activation energy and ${\text{R}}$: Universal gas constant

Complete Step by step solution:
We know that ${{\text{e}}^{\dfrac{{ - {\text{Ea}}}}{{{\text{RT}}}}}}$ in 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${{\text{E}}_{\text{a}}}$. This equation is used to study the dependence of a reaction on temperature.
As we know, the Arrhenius equation is given as follows:
${\text{k = A}}{{\text{e}}^{\dfrac{{ - {\text{Ea}}}}{{{\text{RT}}}}}}$ , taking log on both sides:
$\Rightarrow {\text{lnk}} = {\text{lnA}} - {{\text{E}}_{\text{a}}}{\text{/RT}}$.
Now compare this with the general equation of a straight line ${\text{y = mx + c}}$.
The slope of ${\text{lnK vs }}\dfrac{1}{{\text{T}}}$ can be written as
$m =$ ${\text{tan}}{45^o}$=$\dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}} = 1$
We know that value of Universal gas constant is ${\text{R = }}2{\text{ cal}}$
$\therefore {{\text{E}}_{\text{a}}} = {\text{ R = }}2{\text{ cal}}$​=$R$=2 Cal
Therefore, we can conclude that the correct answer to this question is option A.

Additional information:
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.

Note: You should know the importance of the Arrhenius equation. You can recall that ${\text{RT}}$ is the average kinetic energy, it becomes apparent that the exponent is just the ratio of the activation energy ${{\text{E}}_{\text{a}}}$ to the average kinetic energy. The 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.