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?

• A. $E_a>E_c>E_d>E_b$
• B. $E_c>E_a>E_d>E_b$
• C. $E_b>E_d>E_c>E_a$
• D. $E_b>E_a>E_d>E_c$

Answer 1

The correct option is B $E_c>E_a>E_d>E_b$

To avoid confusion, in this question we’ll be 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 .............(2)
Comparing equation 1 with 2 we get,
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.
$E_c>E_a>E_d>E_b$