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:

Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}.$ If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300 \, K,$ then $\ln (k_2/k_1)$ is equal to :
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$

$A$ catalyst:

The value of the rate constant for the reaction $A + B \to \text{products}$ depends on:

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 quadruples when the temperature changes from $300 \, K$ to $310 \, K$. The activation energy of this reaction is ........... $kJ \, mol^{-1}$ (Assume activation energy and pre-exponential factor are independent of temperature; $\ln 2 = 0.693$; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$)