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

The activation energy of a reaction at a given temperature is found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of rate constant to the Arrhenius factor is

How fast is the reaction at $25\, ^\circ C$ as compared to $0\, ^\circ C$,if the activation energy is $65\, kJ/mol$?

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 $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

For reaction $A \to B$,the rate constant $k_1 = A_1 e^{-E_{a_1} / (RT)}$ and for the reaction $X \to Y$ the rate constant $k_2 = A_2 e^{-E_{a_2} / (RT)}$. If $A_1 = 10^8$,$A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal/mol$,$E_{a_2} = 1800 \ cal/mol$,then the temperature at which $k_1 = k_2$ is (Given : $R = 2 \ cal/K \cdot mol$)

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