When the temperature changes from $20\,^oC$ to $50\,^oC$,the rate of reaction becomes three times. The activation energy $(E_a)$ for the reaction is $.... \, kJ \, mol^{-1}$ $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

The rate of the chemical reaction doubles for an increase of $10 \, K$ in absolute temperature from $298 \, K$. Calculate $E_{a}$.

For a reaction taking place in three steps at the same temperature,the overall rate constant is $K = \frac{K_1 K_2}{K_3}$. If $E_{a1}$,$E_{a2}$,and $E_{a3}$ are $40 \ kJ/mol$,$50 \ kJ/mol$,and $60 \ kJ/mol$ respectively,the overall activation energy $E_a$ is . . . . . . $kJ/mol$.

$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)

The number of molecules with energy greater than the threshold energy for a reaction increases five-fold by a rise of temperature from $27^{\circ} C$ to $42^{\circ} C$. Its energy of activation in $J \ mol^{-1}$ is.............. (Take $\ln 5 = 1.6094$; $R = 8.314 \ J \ mol^{-1} K^{-1}$)