Chemical Kinetics question bank (Chemistry)

$\mathrm{A} \longrightarrow \mathrm{B}$ (first reaction)
$\mathrm{C} \longrightarrow \mathrm{D}$ (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 $500 \mathrm{~K}, 50 \%$ 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 $\_\_\_\_$ $\times 10^{-1}$ hour $^{-1}$ (nearest integer).

Reactant A converts to product D through the given mechanism (with the net evolution of heat) :
$\mathrm{A} \rightarrow \mathrm{B} \quad$ slow $; \Delta \mathrm{H}=+\mathrm{ve}$
$\mathrm{B} \rightarrow \mathrm{C}$ fast; $\Delta \mathrm{H}=-\mathrm{ve}$
$\mathrm{C} \rightarrow \mathrm{D} \quad$ fast ; $\Delta \mathrm{H}=-\mathrm{ve}$
Which of the following represents the above reaction mechanism?