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

The energies of activation for forward and reverse reactions for $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \, kJ \, mol^{-1}$ and $200 \, kJ \, mol^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \, kJ \, mol^{-1}.$ The enthalpy change of the reaction $(A_2 + B_2 \rightarrow 2AB)$ in the presence of a catalyst will be (in $kJ \, mol^{-1}$)

Which of the following statements is correct?

For the decomposition reaction of $N_2O_5$,the slope of the graph of $\log K$ versus $1/T$ is $-1.2 \times 10^4 \ K$. Calculate the activation energy $(E_a)$ of the reaction.

Rate constant varies with temperature by the equation $log_{10} K = 5 - 2000 / T$. We can conclude that $(R = 8.314 \ J \ mol^{-1} K^{-1})$

Consider the following statements:
$(i)$ Increase in concentration of reactant increases the rate of a zero order reaction
$(ii)$ Rate constant $k$ is equal to $A$ if $E_a = 0$
$(iii)$ Rate constant $k$ is equal to $A$ if $E_a = \infty$
$(iv)$ $\log_e k$ vs $T$ is a straight line
$(v)$ $\log_e k$ vs $1/T$ is a straight line
Which of the following statements are correct?