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 $T$?

$A$ reaction takes place in three steps with individual rate constants and activation energies. The overall rate constant is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$. The overall activation energy of the reaction in $kJ/mol$ is:

$Step$	$Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$	$k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$	$k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$	$k_3, E_{a_3} = 50\ kJ/mol$

$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
The curves $M$ and $N$ represent the variation of energy with reaction coordinate for the reaction in absence and presence of catalyst.
Which value represents the activation energy $(E_a)$ for the backward reaction in the presence of catalyst?

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.

Explain the rate of reaction with energy of activation and temperature with the help of the Arrhenius equation and state its importance.

Give an example of a collision in the proper direction that results in the formation of a product.