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 $kcal \ mol^{-1}$,$R=2 \ cal \ mol^{-1} \ K^{-1}$).

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the activation energies $(E_a)$ for the forward and backward reactions are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$ respectively. If a catalyst lowers the $E_a$ for both reactions by $100 \ kJ \ mol^{-1}$,which of the following statements is correct?

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?

The correct reaction profile diagram for a positive catalyst reaction.

The equilibrium constant at $27\,^{\circ}C$ and $127\,^{\circ}C$ is $30$ and $40$ respectively. What will be the ratio of activation energy $(E_{a})_{f} / (E_{a})_{b}$?