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 $kJ \,mol^{-1}$? $(R = 8.314 \,J \,mol^{-1} \,K^{-1})$

The energy of activation is

The temperature coefficient of the saponification reaction of an ester by $NaOH$ is $1.75$. The activation energy of the reaction is .......... $kcal \ mol^{-1}$. (in $.21$)

If for two reactions $E_{a_1} > E_{a_2}$ and $\mu_1$ and $\mu_2$ are their respective temperature coefficients,then which of the following is correct?

For a first-order reaction $A \rightarrow P$,the rate constant $k$ depends on temperature $(T)$ according to the equation $\log \, k = - \frac{2000}{T} + 0.6$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

The $\Delta H$ value of the reaction $H_2 + Cl_2 \rightleftharpoons 2HCl$ is $-44.12 \ kcal$. If $E_1$ is the activation energy of the backward reaction and $E_2$ is the activation energy of the forward reaction,then for the above reaction: