Sol. $k=A \cdot e^{-E a / R T}$ $$ \log _y k=\log _c A-\underbrace{\frac{E a}{2.303 R}}_m \cdot \frac{1}{T} $$ Higher is Ea, stronger is the temperature dependence of k (i.e. steeper the slope) $$ \begin{aligned} (B) & \Rightarrow \frac{1}{k} \frac{d k}{d T}=\frac{E a}{R} \frac{1}{T^2} \\ & \Rightarrow \frac{d k}{d T}=A \times e^{-\frac{E a}{R}} \cdot \frac{E a}{R T^2} \end{aligned} $$