Energy of activation of a reactant is reduced by

In a first order reaction at $27\,^oC$ and $47\,^oC$,$50\%$ of the reaction is complete in $30\,\min$ and $10\,\min$ respectively. Calculate the energy of activation $(E_a)$.

For the reaction $X \rightleftharpoons Y$,the pre-exponential factors for the forward and backward reactions are equal. What will be the equilibrium constant of the reaction?

$A$ reaction is catalysed by $X$. Here $X$:

The rate of a chemical reaction doubles with every $10^{\circ}C$ rise in temperature. If the reaction is carried out in the vicinity of $22^{\circ}C$,the activation energy of the reaction is (Given $R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\ln 2 = 0.69$ and $\ln 3 = 1.1$)

The rate of a reaction doubles when its temperature changes from $300 \, K$ to $310 \, K.$ Activation energy of such a reaction will be .......... $kJ \, mol^{-1}$. $(R= 8.314 \, J \, K^{-1} \, mol^{-1}$ and $\log 2=0.301)$