According to Arrhenius equation, the graph of $\log \mathrm{k}$ vs $\left(\frac{1}{\mathrm{T}}\right)$ has slope equal to:

The reactant $\left(\mathrm{A}\right)$ forms two products:

$\mathrm{A}\stackrel{{\mathrm{k}}_1}{⟶}\mathrm{B}$: Activation energy $={\mathrm{E}}_{{\mathrm{a}}_1}$

$\mathrm{A}\stackrel{{\mathrm{k}}_2}{⟶}\mathrm{C}$: Activation energy $={\mathrm{E}}_{{\mathrm{a}}_2}$

If ${\mathrm{E}}_{{\mathrm{a}}_2}=2{\mathrm{E}}_{{\mathrm{a}}_1},$ then ${\mathrm{k}}_1$ and ${\mathrm{k}}_2$ will be related as:

Assume that for both the reactions, the value of Arrhenius constant is the same.

The accompanying figure depicts the change in concentration of species $\mathrm{X}$ and $\mathrm{Y}$ for the elementary reaction $\mathrm{X}\to \mathrm{Y}$ as a function of time. The point of intersection of two curves represents:

What will be the order of reaction for a chemical change having the graph between $\log {\mathrm{t}}_{1/2}$ vs $\log \mathrm{a}$ as shown below? $($where, $\mathrm{a}=$initial concentration of reactant; $\left.{\mathrm{t}}_{1/2}=\mathrm{half}-\mathrm{life}\right)$