For a reaction, consider the plot of $lnk$ versus $1/T$ given in the figure. If the rate constant of this reaction at $400K$ is ${10}^{-5}{s}^{-1}$ , then the rate constant at $500K$ is:

In the following reaction; $\mathrm{x}\mathrm{A}\to \mathrm{y}\mathrm{B}$

${\log }_{10}\left[-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{d}\mathrm{t}}\right]={\log }_{10}\left[-\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{d}\mathrm{t}}\right]+0.3010$

‘A’ and ‘B’ respectively can be:

For an elementary chemical reaction, ${\mathrm{A}}_2 \underset{{\mathrm{k}}_{-1}}{\overset{{\mathrm{k}}_1}{\rightleftarrows }}2\mathrm{A}$ , the expression for $\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$ is:

Consider the given plots for a reaction obeying Arrhenius equation $(0°C<T<300°C):$ ($K$and $E_a$ are rate constant and activation energy, respectively )

(I) 

(II)