Question

Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?


<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> JEE MAINS 2020

Answer 1

Denoting activation energy by $E_x$
$K=A{e}^{\frac{-E_x}{RT}}$
$\text{log K = log A}-\frac{E_x}{2.303RT}.....\left(1\right)$
Here, the graph given in the question is of a straight line and we know that the equation of straight line is
$y=mx+c....\left(2\right)$
Comparing equation 1 with 2 we get,
$\text{Slope}=\frac{-E_x}{2.303R}$
So, from the graph we can conclude that the line with the most negative slope will have the maximum activation energy value.
The correct order will be
$E_c>E_a>E_d>E_b$