Which of the following graphs for $\ln k$ versus $\frac{1}{T}$ is correct?

The rate constant of a reaction varies with temperature according to the equation: $\log K = \text{constant} - \frac{E_a}{2.303 RT}$. If a plot of $\log K$ versus $1/T$ yields a straight line with a slope of $-5632$,then the activation energy of the reaction is .......... $kJ \ mol^{-1}$.

The number of correct statement/s from the following is:
$A.$ Larger the activation energy,smaller is the value of the rate constant.
$B.$ The higher is the activation energy,higher is the value of the temperature coefficient.
$C.$ At lower temperatures,the increase in temperature causes a larger change in the value of $k$ than at higher temperatures.
$D.$ $A$ plot of $\ln k$ vs $\frac{1}{T}$ is a straight line with a slope equal to $-\frac{E_a}{R}$.

The half-life periods of a first order reaction at $300 \ K$ and $400 \ K$ are $50 \ s$ and $10 \ s$ respectively. The activation energy of the reaction in $kJ \ mol^{-1}$ is $(\log 5 = 0.70)$

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)

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