The minimum energy necessary to permit a reaction is

The activation energy of a first-order reaction at $25\,^{\circ}C$ is $30\,kJ/mol$. In the presence of a catalyst,the activation energy of the same reaction at $25\,^{\circ}C$ becomes $24\,kJ/mol$. The rate of the reaction in the presence of the catalyst will be how many times the original rate?

What does $P \cdot Z_{AB} \cdot e^{-\frac{E_a}{RT}}$ indicate in the rate equation?

Consider the following reaction that proceeds from $A$ to $B$ in three steps as shown in the energy profile diagram. Choose the correct values for the following parameters:
$1$. Number of intermediates
$2$. Number of activated complexes
$3$. Rate determining step

The energies of activation for forward and reverse reactions for $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \ kJ \ mol^{-1}$. The enthalpy change of the reaction $(A_2 + B_2 \to 2AB)$ in the presence of catalyst will be (in $kJ \ mol^{-1}$)

$A \rightarrow B$ (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).