In the presence of a catalyst,the heat evolved or absorbed during the reaction . . . . . . .

For a certain gaseous reaction,the temperature is increased by $10\,^{\circ}C$ from $25\,^{\circ}C$ to $35\,^{\circ}C$. If the rate of the reaction doubles,what will be the value of the activation energy $(E_a)$?

The rate constant of a reaction at temperature $200 \ K$ is $10$ times less than the rate constant at $400 \ K.$ What is the activation energy $({E_a})$ of the reaction ($R = $ gas constant) (in $R$)?

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?

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:

What will be the effect of temperature on the rate constant?