$A$ reaction takes place in three steps with individual rate constants and activation energies. The overall rate constant is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$. The overall activation energy of the reaction in $kJ/mol$ is:

$Step$	$Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$	$k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$	$k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$	$k_3, E_{a_3} = 50\ kJ/mol$

$A \rightarrow B$ (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).

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

$A$ catalyst:

For a reaction at a given temperature $T$,the rate constant $K$ is given by the Arrhenius equation. What is the value of $\lim_{T \to \infty} \log K$?

The specific rate constant of a first-order reaction depends on .......