For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is $6.64 \ kJ \ mol^{-1}$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$, the temperature at which the rate constant becomes $e^2 \ min^{-1}$, is (in $K$)

Which of the following statements is correct regarding the Arrhenius equation and reaction kinetics?

Correct statements regarding Arrhenius equation among the following are:
$(A)$ Factor $e^{-Ea/RT}$ corresponds to fraction of molecules having kinetic energy less than $Ea$.
$(B)$ At a given temperature, lower the $Ea$, faster is the reaction.
$(C)$ Increase in temperature by about $10^{\circ}C$ doubles the rate of reaction.
$(D)$ Plot of $\log k$ vs $\frac{1}{T}$ gives a straight line with $slope = -\frac{Ea}{2.303R}$.
Choose the correct answer from the options given below:

The activation energy of a chemical reaction can be determined by:

On introducing a catalyst at $500 \, K,$ the rate constant of a first order reaction increases $2.718$ times. If the activation energy in the presence of a catalyst is $4.15 \, kJ \, mol^{-1},$ then what will be $E_a$ in the absence of a catalyst? (Value of $e = 2.718$ and $R = 8.314 \times 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$)

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.