How fast is the reaction at $25\, ^\circ C$ as compared to $0\, ^\circ C$,if the activation energy is $65\, kJ/mol$?

What do $Z_{AB}$ and $P$ indicate in the context of collision theory?

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.

For a reaction,the rate constant is expressed as,$K = A \cdot e^{-40000/T}$. The energy of activation is ....... $cal$.

Consider $A \xrightarrow{k_1} B$ and $C \xrightarrow{k_2} D$ are two reactions. If the rate constant $(k_1)$ of the $A \rightarrow B$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T/K}$ and activation energy of $C \rightarrow D$ reaction $(Ea_2)$ is $\frac{1}{5}$th of the $A \rightarrow B$ reaction $(Ea_1)$, then the value of $(Ea_2)$ is . . . . . . $kJ \ mol^{-1}$. (Nearest Integer)

For the equilibrium $A_{(g)} \rightleftharpoons B_{(g)}$,$\Delta H$ is $-40 \ kJ/mol$. If the ratio of the activation energies of the forward $(E_f)$ and reverse $(E_b)$ reactions is $\frac{2}{3}$,then: