A → B (first reaction)

C → D (second reaction)

Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is _______ × 10^-1 hour^-1 (nearest integer).

The half-life of ${ }^{65} \mathrm{Zn}$ is 245 days. After $x$ days, $75 \%$ of original activity remained. The value of $x$ in days is $\_\_\_\_$ . (Nearest integer)

(Given: $\log 3=0.4771$ and $\log 2=0.3010$ )

For the thermal decomposition of reactant $\mathrm{AB}(\mathrm{g})$, the following plot is constructed.

The half life of the reaction is ' $x^{\prime} \,\mathrm{min}$.

$x=$ $\_\_\_\_$ min. (Nearest integer)

Consider $\mathrm{A} \xrightarrow{\mathrm{k}_1} \mathrm{~B}$ and $\mathrm{C} \xrightarrow{\mathrm{k}_2} \mathrm{D}$ are two reactions. If the rate constant $\left(\mathrm{k}_1\right)$ of the $\mathrm{A} \longrightarrow \mathrm{B}$ reaction can be expressed by the following equation $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}}$ and activation energy of $C \longrightarrow D$ reaction $\left(E a_2\right)$ is $\frac{1}{5}$ th of the $A \longrightarrow B$ reaction $\left(E a_1\right)$, then the value of $\left(E a_2\right)$ is

$\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)