The graph obtained between $\ln k$ ($k=$ Rate constant) on $y$-axis and $1/T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$

$A \rightarrow B$ (first reaction)
$C \rightarrow D$ (second reaction)
Consider the above two first-order reactions. The rate constant for the first reaction at $500 \ K$ is double of the same at $300 \ K$. At $500 \ K, 50 \%$ of the reaction becomes complete in $2 \ hours$. The activation energy of the second reaction is half of that of the first reaction. If the rate constant at $500 \ K$ of the second reaction is double the rate constant of the first reaction at the same temperature,then the rate constant for the second reaction at $300 \ K$ is . . . . . . $\times 10^{-1} \ hour^{-1}$ (nearest integer).

$A$ catalyst:

Consider a complex reaction taking place in three steps with rate constants $k_1$,$k_2$,and $k_3$ respectively. The overall rate constant $k$ is given by the expression $k = \sqrt{\frac{k_1 k_3}{k_2}}$. If the activation energies of the three steps are $60$,$30$,and $10 \ kJ \ mol^{-1}$ respectively,then the overall energy of activation in $kJ \ mol^{-1}$ is $..........$ $(Nearest \ integer)$

The rate constant for a first order reaction is $0.58 \ s^{-1}$ at $300 \ K$ and $0.026 \ s^{-1}$ at $290 \ K$. What is the energy of activation? $(R=8.314 \ J \ K^{-1} \ mol^{-1})$

For a reaction,activation energy $E_a = 0$ and rate constant $K = 3.2 \times 10^6 \ s^{-1}$ at $300 \ K$. What is the value of the rate constant at $300 \ K$?