The Arrhenius equation expressing the effect of temperature on the rate constant of a reaction is

What is the activation energy $(kJ \, mol^{-1})$ for a reaction if its rate constant doubles when the temperature is raised from $300 \, K$ to $400 \, K$ ? $(R = 8.314 \, J \, mol^{-1} \, K^{-1})$

$A \rightarrow B$ (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).

If for a hypothetical reaction,$E_a = 0$ at $273 \ K$,then find the ratio of the rate constants at $383 \ K$ and $273 \ K$.

An endothermic reaction $A \to B$ has an activation energy of $15 \ kcal/mol$ and an enthalpy change of reaction of $5 \ kcal/mol$. The activation energy of the reverse reaction $B \to A$ is ......... $kcal/mol$.

Write the logarithmic form of the Arrhenius equation $k = A e^{-\frac{E_a}{RT}}$.