For an endothermic reaction $X \rightarrow Y$,the activation energies for the forward and backward reactions are $E_f$ and $E_b$ respectively. Then,in general:

$A$ large increase in the rate of a reaction for a rise in temperature is due to:

The activation energy for the reaction $X \rightarrow Y$ is $150 \text{ kJ mol}^{-1}$. The change in enthalpy for the above reaction is $-135 \text{ kJ mol}^{-1}$. What is the activation energy for the reverse reaction $Y \rightarrow X$?

What is the activation energy $(kJ \, mol^{-1})$ for a reaction if its rate constant doubles when the temperature is raised from $300 \, K$ to $400 \, K$ ? $(R = 8.314 \, J \, mol^{-1} \, K^{-1})$

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

Write the Arrhenius equation in the form $\ln \, k = -\frac{E_a}{RT} + \ln \, A$.