Activation energy of any reaction depends on

The Arrhenius equation is represented as $k = A e^{-E_a/RT}$. The activation energy $E_a$ of the reaction can be calculated by plotting:

At temperature $T \ K$,the rate constant of a reaction is $1/10$th of the rate constant at temperature $2T \ K$. What will be the activation energy of this reaction (in $RT$)?

For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

$A$ chemical reaction was carried out at $300 \, K$ and $280 \, K$. The rate constants were found to be $K_1$ and $K_2$ respectively. The energy of activation is $1.157 \times 10^4 \, cal \, mol^{-1}$ and $R = 1.987 \, cal \, K^{-1} \, mol^{-1}$. Then:

For a complex reaction $A \xrightarrow{K} \text{products}$,where $Ea_1 = 180 \ kJ/mol$,$Ea_2 = 80 \ kJ/mol$,and $Ea_3 = 50 \ kJ/mol$,the overall rate constant $K$ is related to individual rate constants by the equation $K = (\frac{K_1 \cdot K_2}{K_3})^{2/3}$. The activation energy $(kJ/mol)$ for the overall reaction is: