The time taken for half of the initial amount of $N_2O_5$ to decompose is $12 \ min$ at $310 \ K$ and $2 \ hrs$ at $300 \ K$. The activation energy of the reaction in $kJ \ mol^{-1}$ is $\left(R=8.3 \ J \ K^{-1} \ mol^{-1}\right)$

For a first order reaction,a plot of $\ln k$ ($y$-axis) and $\frac{1}{T}$ ($x$-axis) gave a straight line with a slope equal to $-10^3 \ K$ and an intercept equal to $2.303$ (on $y$-axis). What is the activation energy ($E_a$ in $kJ \ mol^{-1}$) of the reaction? (Given $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)

$A$ 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:

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20 \,^{\circ}C$ to $35 \,^{\circ}C$ in $kJ \,mol^{-1}$? $(R = 8.314 \,J \,mol^{-1} \,K^{-1})$

For a particular reaction,the rate constant becomes double on increasing temperature from $27^{\circ} C$ to $37^{\circ} C$. Calculate the approximate activation energy (in $kcal \ mol^{-1}$,$R=2 \ cal \ mol^{-1} \ K^{-1}$).

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.
[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]