The time taken for half of the initial amount of $N_2O_5$ to decompose is $12 \ min$ at $310 \ K$ and $2 \ hrs$ at $300 \ K$. The activation energy of the reaction in $kJ \ mol^{-1}$ is $\left(R=8.3 \ J \ K^{-1} \ mol^{-1}\right)$

Consider the following transformation involving first order elementary reactions in each step at constant temperature as shown below.
$A + B \underset{\text{Step } 3}{\overset{\text{Step } 1}{\rightleftharpoons}} C \xrightarrow{\text{Step } 2} P$
Some details of the above reaction are listed below.

Step	Rate constant $(s^{-1})$	Activation energy $(kJ \ mol^{-1})$
$1$	$k_1$	$300$
$2$	$k_2$	$200$
$3$	$k_3$	$Ea_3$

If the overall rate constant of the above transformation $(k)$ is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy $(E_a)$ is $400 \ kJ \ mol^{-1}$,then the value of $Ea_3$ is $\qquad$ $kJ \ mol^{-1}$ (nearest integer).

For the reaction of $H_2$ with $I_2$,the rate constant is $2.5 \times 10^{-4} \ dm^3 \ mol^{-1} \ s^{-1}$ at $327 \ ^oC$ and $1.0 \ dm^3 \ mol^{-1} \ s^{-1}$ at $527 \ ^oC$. The activation energy for the reaction,in $kJ \ mol^{-1}$ is: $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

$A$ reaction takes place in three steps with individual rate constant and activation energy,

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

Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.

What is the relationship between the temperature $T$ and the activation energy $E_a$ in the Arrhenius equation?

Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.
The correct answer is