If the rate of a reaction doubles when the temperature increases from $298 \, K$ to $308 \, K$,the activation energy of the reaction is ........... $kJ \, mol^{-1}$.

How does the graph of the fraction of molecules versus kinetic energy change at different temperatures?

Which of the following statements is correct for the activation energy of a reaction?

For the reaction,following data is given,
$A \rightarrow B$; $K_1 = 10^{15} \exp \left( \frac{-2000}{T} \right)$
$C \rightarrow D$; $K_2 = 10^{14} \exp \left( \frac{-1000}{T} \right)$
The temperature at which $K_1 = K_2$ is ........... $K$ $(exp. = e)$

The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

The variation of the rate constant with temperature is given by the Arrhenius equation $k = A e^{-E_a / (RT)}$. If $T \to \infty$,the rate constant $k$ will be equal to: