A graph is plotted between log $\mathrm{K}$ versus $\frac{1}{\mathrm{T}}$ for calculation of activation energy $\left({\mathrm{E}}_{\mathrm{a}}\right)$. The correct plot is

According to Arrhenius equation,

$\mathrm{k}\propto \mathrm{T}$ and $\mathrm{k}={\mathrm{Ae}}^{-\mathrm{Ea}/\mathrm{RT}}$

$\mathrm{logk}=\mathrm{logA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times \mathrm{R}}\times \frac{1}{\mathrm{T}}$

$\mathrm{y}=\mathrm{mx}+\mathrm{c}$, the equation of straight line

The value of log $\mathrm{k}$ should increase uniformily with $\mathrm{T}$ or decrease with $\frac{1}{\mathrm{T}}$.

From this, that the slope is negative, which means that angle with x - axis is greater than 90° also that the point on which the straight line of graph intercepts the y - axis is log A which is positive.

Hence all the conclusions point out towards option b as it has negative slope and a positive y intercept.